# $\lim_{n \to \infty} \sum_{k=1}^n\frac 1 { \sqrt{n^2+k} }$

(1) $$\lim_{n \to \infty} \sum_{k=1}^n \frac 1 { \sqrt{n^2+k} }$$

(2) $$\lim_{n\to\infty} \frac {1+\sqrt[n]2 + \sqrt[n]3 + ... \sqrt[n]n} {n}$$

The answers should both be 1.. any hints?

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(2) is a duplicate of math.stackexchange.com/questions/592581/… – Hagen von Eitzen Dec 9 '13 at 22:12
Hint : for the second see my answer here math.stackexchange.com/questions/561422/… – Haha Dec 9 '13 at 22:14
The sum in (1) does not converge for any $n$, hence the limit does not exist – Hagen von Eitzen Dec 9 '13 at 22:14
For the first, you may mean the sum from $1$ to $n$. – André Nicolas Dec 9 '13 at 22:33

For the first problem, note that if $1\le k\le n$, then $$n^2\lt n^2+k\le n^2+n\lt\left(n+\frac{1}{2}\right)^2.$$ Thus $$\frac{1}{n+\frac{1}{2}}\le \frac{1}{\sqrt{n^2+k}}\lt \frac{1}{n}.$$ Our sum is therefore between $\frac{n}{n+\frac{1}{2}}$ and $1$.

Squeeze.

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Here is an alternative proof for (2) that is not in the suggested answers (without employing Cezaro-Stolz).

$$\lim_{n \rightarrow \infty} \frac{1+(2)^{1/n} + ... + (n)^{1/n}}{n} \\ = \lim_{n \rightarrow \infty} \frac{n^{1/n}\cdot ((1/n)^{1/n}+(2/n)^{1/n} + ... + (n/n)^{1/n})}{n} \\ = \lim_{n \rightarrow \infty} \frac{((1/n)^{1/n}+(2/n)^{1/n} + ... + (n/n)^{1/n})}{n} \\ = \lim_{m,n \rightarrow \infty} \frac{((1/n)^{1/m}+(2/n)^{1/m} + ... + (n/n)^{1/m})}{n} \\ = \lim_{m \rightarrow \infty} \int_{0}^{1} x^{1/m}\; dx \\ = \lim \frac{1^{1+1/m}}{1+1/m} \\ = 1$$

NOTE: that $\lim n^{1/n} = \exp(\lim \frac{\log(n)}{n}) = \exp(0) = 1$ and so the second step is allowed. Also, you probably want to tighten up the analysis regarding uniform convergence though.

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