2
$\begingroup$

Find the following limit $$\lim_{n \rightarrow \infty} (\frac{2^{1/n}}{n+1} +\frac{2^{2/n}}{n+1/2}+\dots+\frac{2^{n/n}}{n+{1/n}})$$

My attempts : $$\frac{2^{1/n}}{n+1} +\frac{2^{2/n}}{n+1/2}+....+\frac{2^{n/n}}{n+{1/n}}= \sum_{k=1}^{n} \frac{2^{k/n}}{n+1/k} =\frac{1}{n} \sum_{k=1}^{n} \frac{2^{k/n}}{1+1/kn}.$$ After that I was thinking about Riemann sum, but $2^{k/n}$ creates confusion in my mind. So I'm not able to proceed further.

Any hints/solution will be appreciated. Thanks.

$\endgroup$

2 Answers 2

2
$\begingroup$

Hint. Note that $$\frac{1}{1+\frac{1}{n}}\cdot\frac{1}{n} \sum_{k=1}^{n} 2^{k/n}\leq \sum_{k=1}^{n} \frac{2^{k/n}}{n+\frac{1}{k}}\leq \frac{1}{n} \sum_{k=1}^{n} 2^{k/n}.$$ Then use the fact that $\frac{1}{n} \sum_{k=1}^{n} 2^{k/n}$ is a Riemann sum of $\int_0^1 2^x\,dx$.

$\endgroup$
2
  • $\begingroup$ Give me 5 minutes let me thinks about ur answer $\endgroup$
    – jasmine
    Nov 7, 2018 at 17:38
  • $\begingroup$ thanks U @Robert $\endgroup$
    – jasmine
    Nov 7, 2018 at 17:42
1
$\begingroup$

It is a Riemann sum as you say: $\int_0^1 2^x \, dx = 1/\ln 2$. The denominator vanishes as $n\to\infty$.

$\endgroup$
2
  • $\begingroup$ I don't think the given sum is Riemann sum of the function you're using. As shown in the other answer, it though can be sandwiched between two expressions using that Riemann sum... $\endgroup$
    – DonAntonio
    Nov 7, 2018 at 17:42
  • $\begingroup$ Agreed, it isn't, but the difference is zero in the limit as you say. I think we were typing away at the same time ... $\endgroup$ Nov 7, 2018 at 18:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .