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Find the limit of summation $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\dfrac{{{2^{\frac{k}{n}}}}}{{n + \frac{1}{k}}}}.$$

Can I use $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\dfrac{{{2^{\frac{k}{n}}}}}{{n + \frac{1}{k}}}} = \mathop {\lim }\limits_{n \to \infty }\sum\limits_{k = 1}^n {\dfrac{{{2^{\frac{k}{n}}}}}{n}}?$$

Thank you for any help.

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2 Answers 2

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Yes you can but it will be a less than or equal inequality instead. This gives a upperbound which is $\displaystyle \int_{0}^1 2^xdx$. To prove that this is the answer, we write $n+ \dfrac{1}{k} < n + 1= n\left(1+\dfrac{1}{n}\right)\implies L \ge \dfrac{\displaystyle \int_{0}^1 2^xdx}{ \displaystyle \lim_{n \to \infty} \left(1+\dfrac{1}{n}\right)} = \displaystyle \int_{0}^1 2^x dx$

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This doesn't give an explicit calculation of the limit; it only provides an upper bound to prove convergence.

Hint: Use the following: $$\sum_{k=1}^n\frac{2^{k/n}}{n+(1/k)}\leq\sum_{k=1}^n\frac{(2^{1/n})^k}{n}=\frac 1n\sum_{k=1}^n\left(2^{1/n}\right)^k.$$Now use the fact that the right-hand side is a geometric series.

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