Evaluate $ lim limits_{n to infty }frac{1}{n}sqrt[n]{n^5+(n+1)^5+…+(2n)^5}$











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$ lim limits_{n to infty }frac{1}{n}sqrt[n]{n^5+(n+1)^5+...+(2n)^5}$
I have no idea. Please give me some hint.










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    $ lim limits_{n to infty }frac{1}{n}sqrt[n]{n^5+(n+1)^5+...+(2n)^5}$
    I have no idea. Please give me some hint.










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      favorite
      1









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      $ lim limits_{n to infty }frac{1}{n}sqrt[n]{n^5+(n+1)^5+...+(2n)^5}$
      I have no idea. Please give me some hint.










      share|cite|improve this question















      $ lim limits_{n to infty }frac{1}{n}sqrt[n]{n^5+(n+1)^5+...+(2n)^5}$
      I have no idea. Please give me some hint.







      real-analysis limits limits-without-lhopital






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      edited Nov 17 at 11:34









      Paramanand Singh

      48.4k555156




      48.4k555156










      asked Nov 17 at 10:35









      matematiccc

      1145




      1145






















          3 Answers
          3






          active

          oldest

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          up vote
          3
          down vote



          accepted










          Hint: $$0leq n^5+(n+1)^5+dots+(2n)^5leq (n+1)cdot2n^5leq2ncdot (2n)^5=(2n)^6$$






          share|cite|improve this answer





















          • Thanks! And limit of (n^6)^(1/n) is 1?
            – matematiccc
            Nov 17 at 10:55










          • yes, it's 1 :).
            – b00n heT
            Nov 17 at 10:57










          • $1leq n^5+...$
            – user90369
            Nov 17 at 18:20


















          up vote
          0
          down vote














          1. Use
            $$
            a^x = exp( x cdot ln(a)) forall x in mathbb{R} forall a > 0.
            $$


          2. Then use that
            $$
            lim_{n to infty} e^{f(n)} = exp(lim_{n to infty} f(n)),
            $$

            since $x mapsto e^x$ ist continuous on $mathbb{R}$.


          3. Lastly notice that the limit of a product is the product of the limits when both limits don't approach 0.



          Alternative



          Knowing (looking up...) $sum_{k = 0}^{n} k^5 = frac{n^2 (n + 1)^2 (2 n^2 + 2 n - 1)}{12}$ one may derive
          $$
          sum_{k = n}^{2n} k^5 = frac{n^2 (n + 1) (42 n^3 + 24 n^2 + n - 1)}{4}
          le frac{n^2 cdot n cdot 42n^3}{4} le 11 n^6,
          $$

          where the latter inequalities will be true for large $n$, and proceed in a similar fashion from there






          share|cite|improve this answer






























            up vote
            0
            down vote













            Hint :



            $0 lt (n^6)^{1/n} lt((n+1)n^5)^{1/n} lt $



            $(n^5+(n+1)^5....+(2n)^{5})^{1/n} lt$



            $((n+1)(2n)^5)^{1/n}lt ((2n)^6)^{1/n}.$



            The lower and upper bound limits exist, hence



            $(n^5+(n+1)^5+..(2n)^5)^{1/n} lt M,$



            where $M>0$, real.



            The limit $n rightarrow infty $



            $(1/n)(n^5+(n+1)^5+...(2n)^n)^{1/n}$



            is?






            share|cite|improve this answer























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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              down vote



              accepted










              Hint: $$0leq n^5+(n+1)^5+dots+(2n)^5leq (n+1)cdot2n^5leq2ncdot (2n)^5=(2n)^6$$






              share|cite|improve this answer





















              • Thanks! And limit of (n^6)^(1/n) is 1?
                – matematiccc
                Nov 17 at 10:55










              • yes, it's 1 :).
                – b00n heT
                Nov 17 at 10:57










              • $1leq n^5+...$
                – user90369
                Nov 17 at 18:20















              up vote
              3
              down vote



              accepted










              Hint: $$0leq n^5+(n+1)^5+dots+(2n)^5leq (n+1)cdot2n^5leq2ncdot (2n)^5=(2n)^6$$






              share|cite|improve this answer





















              • Thanks! And limit of (n^6)^(1/n) is 1?
                – matematiccc
                Nov 17 at 10:55










              • yes, it's 1 :).
                – b00n heT
                Nov 17 at 10:57










              • $1leq n^5+...$
                – user90369
                Nov 17 at 18:20













              up vote
              3
              down vote



              accepted







              up vote
              3
              down vote



              accepted






              Hint: $$0leq n^5+(n+1)^5+dots+(2n)^5leq (n+1)cdot2n^5leq2ncdot (2n)^5=(2n)^6$$






              share|cite|improve this answer












              Hint: $$0leq n^5+(n+1)^5+dots+(2n)^5leq (n+1)cdot2n^5leq2ncdot (2n)^5=(2n)^6$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 17 at 10:41









              b00n heT

              10.2k12134




              10.2k12134












              • Thanks! And limit of (n^6)^(1/n) is 1?
                – matematiccc
                Nov 17 at 10:55










              • yes, it's 1 :).
                – b00n heT
                Nov 17 at 10:57










              • $1leq n^5+...$
                – user90369
                Nov 17 at 18:20


















              • Thanks! And limit of (n^6)^(1/n) is 1?
                – matematiccc
                Nov 17 at 10:55










              • yes, it's 1 :).
                – b00n heT
                Nov 17 at 10:57










              • $1leq n^5+...$
                – user90369
                Nov 17 at 18:20
















              Thanks! And limit of (n^6)^(1/n) is 1?
              – matematiccc
              Nov 17 at 10:55




              Thanks! And limit of (n^6)^(1/n) is 1?
              – matematiccc
              Nov 17 at 10:55












              yes, it's 1 :).
              – b00n heT
              Nov 17 at 10:57




              yes, it's 1 :).
              – b00n heT
              Nov 17 at 10:57












              $1leq n^5+...$
              – user90369
              Nov 17 at 18:20




              $1leq n^5+...$
              – user90369
              Nov 17 at 18:20










              up vote
              0
              down vote














              1. Use
                $$
                a^x = exp( x cdot ln(a)) forall x in mathbb{R} forall a > 0.
                $$


              2. Then use that
                $$
                lim_{n to infty} e^{f(n)} = exp(lim_{n to infty} f(n)),
                $$

                since $x mapsto e^x$ ist continuous on $mathbb{R}$.


              3. Lastly notice that the limit of a product is the product of the limits when both limits don't approach 0.



              Alternative



              Knowing (looking up...) $sum_{k = 0}^{n} k^5 = frac{n^2 (n + 1)^2 (2 n^2 + 2 n - 1)}{12}$ one may derive
              $$
              sum_{k = n}^{2n} k^5 = frac{n^2 (n + 1) (42 n^3 + 24 n^2 + n - 1)}{4}
              le frac{n^2 cdot n cdot 42n^3}{4} le 11 n^6,
              $$

              where the latter inequalities will be true for large $n$, and proceed in a similar fashion from there






              share|cite|improve this answer



























                up vote
                0
                down vote














                1. Use
                  $$
                  a^x = exp( x cdot ln(a)) forall x in mathbb{R} forall a > 0.
                  $$


                2. Then use that
                  $$
                  lim_{n to infty} e^{f(n)} = exp(lim_{n to infty} f(n)),
                  $$

                  since $x mapsto e^x$ ist continuous on $mathbb{R}$.


                3. Lastly notice that the limit of a product is the product of the limits when both limits don't approach 0.



                Alternative



                Knowing (looking up...) $sum_{k = 0}^{n} k^5 = frac{n^2 (n + 1)^2 (2 n^2 + 2 n - 1)}{12}$ one may derive
                $$
                sum_{k = n}^{2n} k^5 = frac{n^2 (n + 1) (42 n^3 + 24 n^2 + n - 1)}{4}
                le frac{n^2 cdot n cdot 42n^3}{4} le 11 n^6,
                $$

                where the latter inequalities will be true for large $n$, and proceed in a similar fashion from there






                share|cite|improve this answer

























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote










                  1. Use
                    $$
                    a^x = exp( x cdot ln(a)) forall x in mathbb{R} forall a > 0.
                    $$


                  2. Then use that
                    $$
                    lim_{n to infty} e^{f(n)} = exp(lim_{n to infty} f(n)),
                    $$

                    since $x mapsto e^x$ ist continuous on $mathbb{R}$.


                  3. Lastly notice that the limit of a product is the product of the limits when both limits don't approach 0.



                  Alternative



                  Knowing (looking up...) $sum_{k = 0}^{n} k^5 = frac{n^2 (n + 1)^2 (2 n^2 + 2 n - 1)}{12}$ one may derive
                  $$
                  sum_{k = n}^{2n} k^5 = frac{n^2 (n + 1) (42 n^3 + 24 n^2 + n - 1)}{4}
                  le frac{n^2 cdot n cdot 42n^3}{4} le 11 n^6,
                  $$

                  where the latter inequalities will be true for large $n$, and proceed in a similar fashion from there






                  share|cite|improve this answer















                  1. Use
                    $$
                    a^x = exp( x cdot ln(a)) forall x in mathbb{R} forall a > 0.
                    $$


                  2. Then use that
                    $$
                    lim_{n to infty} e^{f(n)} = exp(lim_{n to infty} f(n)),
                    $$

                    since $x mapsto e^x$ ist continuous on $mathbb{R}$.


                  3. Lastly notice that the limit of a product is the product of the limits when both limits don't approach 0.



                  Alternative



                  Knowing (looking up...) $sum_{k = 0}^{n} k^5 = frac{n^2 (n + 1)^2 (2 n^2 + 2 n - 1)}{12}$ one may derive
                  $$
                  sum_{k = n}^{2n} k^5 = frac{n^2 (n + 1) (42 n^3 + 24 n^2 + n - 1)}{4}
                  le frac{n^2 cdot n cdot 42n^3}{4} le 11 n^6,
                  $$

                  where the latter inequalities will be true for large $n$, and proceed in a similar fashion from there







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 17 at 10:53

























                  answered Nov 17 at 10:47









                  Viktor Glombik

                  489321




                  489321






















                      up vote
                      0
                      down vote













                      Hint :



                      $0 lt (n^6)^{1/n} lt((n+1)n^5)^{1/n} lt $



                      $(n^5+(n+1)^5....+(2n)^{5})^{1/n} lt$



                      $((n+1)(2n)^5)^{1/n}lt ((2n)^6)^{1/n}.$



                      The lower and upper bound limits exist, hence



                      $(n^5+(n+1)^5+..(2n)^5)^{1/n} lt M,$



                      where $M>0$, real.



                      The limit $n rightarrow infty $



                      $(1/n)(n^5+(n+1)^5+...(2n)^n)^{1/n}$



                      is?






                      share|cite|improve this answer



























                        up vote
                        0
                        down vote













                        Hint :



                        $0 lt (n^6)^{1/n} lt((n+1)n^5)^{1/n} lt $



                        $(n^5+(n+1)^5....+(2n)^{5})^{1/n} lt$



                        $((n+1)(2n)^5)^{1/n}lt ((2n)^6)^{1/n}.$



                        The lower and upper bound limits exist, hence



                        $(n^5+(n+1)^5+..(2n)^5)^{1/n} lt M,$



                        where $M>0$, real.



                        The limit $n rightarrow infty $



                        $(1/n)(n^5+(n+1)^5+...(2n)^n)^{1/n}$



                        is?






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Hint :



                          $0 lt (n^6)^{1/n} lt((n+1)n^5)^{1/n} lt $



                          $(n^5+(n+1)^5....+(2n)^{5})^{1/n} lt$



                          $((n+1)(2n)^5)^{1/n}lt ((2n)^6)^{1/n}.$



                          The lower and upper bound limits exist, hence



                          $(n^5+(n+1)^5+..(2n)^5)^{1/n} lt M,$



                          where $M>0$, real.



                          The limit $n rightarrow infty $



                          $(1/n)(n^5+(n+1)^5+...(2n)^n)^{1/n}$



                          is?






                          share|cite|improve this answer














                          Hint :



                          $0 lt (n^6)^{1/n} lt((n+1)n^5)^{1/n} lt $



                          $(n^5+(n+1)^5....+(2n)^{5})^{1/n} lt$



                          $((n+1)(2n)^5)^{1/n}lt ((2n)^6)^{1/n}.$



                          The lower and upper bound limits exist, hence



                          $(n^5+(n+1)^5+..(2n)^5)^{1/n} lt M,$



                          where $M>0$, real.



                          The limit $n rightarrow infty $



                          $(1/n)(n^5+(n+1)^5+...(2n)^n)^{1/n}$



                          is?







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Nov 17 at 12:26

























                          answered Nov 17 at 10:57









                          Peter Szilas

                          10.2k2720




                          10.2k2720






























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