Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Q. Compute the following limit:

$$\lim_{n\rightarrow\infty}\left(\frac{2^{1/n}}{n+1}+\frac{2^{2/n}}{n+1/2}+\cdots+\frac{2^{n/n}}{n+1/n}\right)$$ using integral.

One obvious method is to squeeze it using $$2^{(a-1)/n}<\frac{2^{a/n}}{n+a/n}<2^{a/n}$$ and using intermediate value theorem such that $$\exists c\in({(a-1)/n},a/n) : 2^c=\frac{2^{a/n}}{n+a/n}$$ hence the result. $$\int_0^1 2^x \, dx$$.

Are there any other methods?

share|cite|improve this question
up vote 1 down vote accepted

As is, it is simply a Riemann integral, but because, as $n \rightarrow \infty$, the $n$ in the denominator dominates the $k/n \, \forall k \in \{1,2,\ldots,n\}$, that piece in the denominator may simply be ignored, and the sum in the limit is

$$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n 2^{k/n} = \int_0^1 dx \: 2^x$$

share|cite|improve this answer
OK! cool. and why do we ignore the same quantity in denominator but not in numerator? Merely because we know it should somehow be in the form that is easily reimann integrable? – user45099 Feb 24 '13 at 10:23
No, because it is simply negligible in the limit of $n \rightarrow \infty$ next to $n$. – Ron Gordon Feb 24 '13 at 15:21

Let denote $$S_n=\left(\frac{2^{1/n}}{n+1}+\frac{2^{2/n}}{n+1/2}+...+\frac{2^{n/n}}{n+1/n}\right)=\sum_{k=1}^n\frac{2^{k/n}}{n+1/k}=\frac{1}{n}\sum_{k=1}^n\frac{2^{k/n}}{1+1/nk}.$$ It's obvious that $S_n $ is not a Riemann Sum but we'll show that we can approximate it by the Riemann sum $$T_n=\frac{1}{n}\sum_{k=1}^n2^{k/n},$$ and $\{T_n\}$ converges to $\int_0^1 2^x dx$. Let's now prove that the sequence $\{T_n-S_n\}_n$ converges to $0$ to acheive the proof.

We have $$0\leq T_n-S_n=\frac{1}{n}\sum_{k=1}^n2^{k/n}\frac{1/nk}{1+1/nk}\leq\frac{1}{n}n2^{n/n}\frac{1/n}{1+1/n^2}\leq\frac{2}{n}$$ and by squeeze theorem we conclude.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.