Test if the series $\sum_{n=1}^{\infty} \left(\frac{\sqrt{n}-1}{\sqrt{n}} \right)^n$ converges or diverges I want to test if the series $\sum_{n=1}^{\infty} \left(\frac{\sqrt{n}-1}{\sqrt{n}} \right)^n$ converges or diverges. From my graphing calculator, it looks like each term in this series is nonnegative and is less than the corresponding term in the series $\sum_{n=1}^{\infty} \frac{1}{n^{1.5}}$ So I believe this series converges. But I am wondering if there is a more intuitive proof to show this series converges without using a calculator.
 A: Let $a_n = \left( \frac{\sqrt{n}-1}{\sqrt{n}} \right)^n$. Then for $n \geq 2$, $$\log a_n = n \log \left( 1 - \frac{1}{\sqrt{n}} \right) = -\sqrt{n} + O \left( \frac{1}{\sqrt{n}} \right),$$ which implies $a_n \leq C e^{-\sqrt{n}}$ for some fixed constant $C$ and all $n \geq 1$.
Since $\sum_{n=1}^\infty e^{-\sqrt{n}}$ converges by Limit Comparison Test with $\sum_{n=1}^\infty n^{-2}$, the original series $\sum_{n=1}^\infty a_n$ converges as well.
A: Based on @Greg_Martin's hint. Note that
$$\left(\frac{\sqrt{n}-1}{\sqrt{n}} \right)^n=\left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}$$
Since
$$\lim_{n\to\infty}\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}=\frac{1}{e}<\frac{1}{2}$$
there exists $N\in\mathbb{N}$ such that $n\geq N$ implies
$$\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}<\frac{1}{2}$$
But then for these $n$ we have
$$\left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}<\frac{1}{2^\sqrt{n}}$$
Thus
$$\sum_{n=1}^\infty \left(\frac{\sqrt{n}-1}{\sqrt{n}} \right)^n=\sum_{n=1}^{N-1} \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}+\sum_{n=N}^\infty \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}$$
$$<\sum_{n=1}^{N-1} \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}+\sum_{n=N}^\infty \frac{1}{2^\sqrt{n}}$$
The first sum is a finite sum while the second sum converges by the integral test since
$$\int_N^\infty\frac{1}{2^\sqrt{x}}dx=\frac{\sqrt{N}\ln(2)+1}{2^{\sqrt{N}-1}\ln^2(2)}$$
(see here). We conclude
$$\sum_{n=1}^\infty \left(\frac{\sqrt{n}-1}{\sqrt{n}} \right)^n$$
converges.

EDIT: Here is the last step without the integral test. Note that
$$\lim_{n\to\infty}\frac{n^2}{2^\sqrt{n}}=\lim_{n\to\infty}\exp\left(\ln\left(\frac{n^2}{2^\sqrt{n}}\right)\right)$$
$$=\exp\left(\lim_{n\to\infty}\ln\left(\frac{n^2}{2^\sqrt{n}}\right)\right)=\exp(\lim_{n\to\infty}(2\ln(n)-\sqrt{n}\ln(2)))$$
(by the continuity of the exponential). It is well known that $\ln(n)$ grows slower than any power of $n$. This implies
$$\exp(\lim_{n\to\infty}(2\ln(n)-\sqrt{n}\ln(2)))=\exp(-\infty)=0$$
(to make this step fully rigorous, simply not that if $a_n$ is a sequence such that $a_n\to-\infty$, then $e^{a_n}\to 0$). Thus, there exists $K\in\mathbb{N}$ such that $n\geq K$ implies
$$\frac{n^2}{2^\sqrt{n}}<1$$
Now, define $W=\max\{K,N\}$. Then the sum becomes
$$\sum_{n=1}^{\infty} \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}=\sum_{n=1}^{W-1} \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}+\sum_{n=W}^{\infty} \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}$$
$$<\sum_{n=1}^{W-1} \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}+\sum_{n=W}^{\infty} \frac{1}{2^\sqrt{n}}<\sum_{n=1}^{W-1} \left(\left(1+\frac{(-1)}{\sqrt{n}}\right)^\sqrt{n}\right)^\sqrt{n}+\sum_{n=W}^{\infty} \frac{1}{n^2}$$
The first sum is finite while the second some converges by the p-test.
