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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?

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2 Answers 2

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

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

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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.

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