Prove that $\sum_{k=1}^\infty \left(\frac{\sqrt{k}-1}{\sqrt{k}}\right)^k$ converges using a comparison I have been asked to prove that 
\begin{align*}
\sum_{k=1}^\infty \left(\frac{\sqrt{k}-1}{\sqrt{k}}\right)^k
\end{align*}
converges. I beleive that I was able to do it using a logarithm test with two applications of L'Hopital, but I have been given a hint that it can be done easily with a comparison. However, I am at a loss for which what I can compare it to. Any help would be greatly appreciated.
 A: $\ln(1+x)\le x$ for all $x\ge-1$ so
$$
k^2\left(\frac{\sqrt k-1}{\sqrt k}\right)^k=k^2\left(1-\frac{1}{\sqrt k}\right)^k=k^2\exp\left(k\ln\left(1-\frac{1}{\sqrt k}\right)\right)\le k^2\textrm e^{-\sqrt k}\underset{k\to+\infty}{\to}0,
$$
hence $\sum_{k=1}^{+\infty}\left(\frac{\sqrt k-1}{\sqrt k}\right)^k<+\infty$.
A: Hint $(1-\frac{1}{\sqrt{k}})^{k}=((1-\frac{1}{\sqrt{k}})^{\sqrt{k}})^{\sqrt{k}}\rightarrow e^{-\sqrt{k}}$ 
A: $\begin{array}\\
\dfrac{\sqrt{k}}{\sqrt{k}-1}
&=\dfrac{\sqrt{k}-1+1}{\sqrt{k}-1}\\
&=1+\dfrac{1}{\sqrt{k}-1}\\
\text{so}\\
\left(\dfrac{\sqrt{k}}{\sqrt{k}-1}\right)^k
&=\left(1+\dfrac{1}{\sqrt{k}-1}\right)^k\\
&=\left(\left(1+\dfrac{1}{\sqrt{k}-1}\right)^{\sqrt{k}}\right)^{k/\sqrt{k}}\\
&>e^{k/\sqrt{k}}
\qquad\text{since } (1+\frac1{x-1})^x > e\\
&=e^{\sqrt{k}}\\
&\gt \dfrac{\sqrt{k}^m}{m!}
\qquad\text{for any } m \ge 1\text{ (from power series for } e^x)\\
&\ge \dfrac{k^2}{24}
\qquad\text{choosing } m=4\\
\text{so}\\
\sum_{k=1}^n \left(\frac{\sqrt{k}-1}{\sqrt{k}}\right)^k
&<\sum_{k=1}^n e^{-\sqrt{k}}\\
&<\sum_{k=1}^n \dfrac{24}{k^2}\\
\end{array}
$
and this sum converges.
A: Start by considering the function:
$$T(k) \equiv \left(1-\frac{1}{\sqrt{k}}\right)^k \exp(\sqrt{k})
\quad \quad \quad \text{for all real } k>0.$$
With a bit of work (which I will leave for you), you can show that $T'(k)>0$, so the function $T$ is strictly increasing.  Since $T(k) \rightarrow 1$ this means that the function increases up to this limiting bound (which it never attains), and so for all $k \in \mathbb{N}$ this function is bounded by $T(k) < 1$.  Using this inequality we obtain:
$$\begin{equation} \begin{aligned}
\text{Your Sum} 
= \sum_{k=1}^\infty T(k) \cdot \exp(-\sqrt{k})
< \sum_{k=1}^\infty \exp(-\sqrt{k}).
\end{aligned} \end{equation}$$
Now, since $\exp(-\sqrt{k})$ is a decreasing function, for all $k \in \mathbb{N}$ we have:
$$\begin{equation} \begin{aligned}
\exp(-\sqrt{k}) = \int \limits_{k-1}^{k} \exp(-\sqrt{k}) dr < \int \limits_{k-1}^{k} \exp(-\sqrt{r}) dr.
\end{aligned} \end{equation}$$
We therefore have:
$$\begin{equation} \begin{aligned}
\text{Your Sum} 
< \sum_{k=1}^\infty \exp(-\sqrt{k})
&< \sum_{k=0}^\infty \int \limits_{k}^{k+1} \exp(-\sqrt{r}) dr \\[6pt]
&= \int \limits_0^\infty \exp(-\sqrt{r}) dr \\[6pt]
&= \Bigg[ - 2(1+\sqrt{r}) \exp(-\sqrt{r})  \Bigg]_{r=0}^{r \rightarrow \infty} \\[6pt]
&= \Bigg[ 0 - ( - 2 ) \Bigg] = 2 < \infty. \\[6pt]
\end{aligned} \end{equation}$$
A: Let's use $1+x\le e^x$. 
Then 
$$\frac{\sqrt{k}-1}{\sqrt{k}} = 1- \frac{1}{\sqrt{k}} \le e^{-\frac{1}{\sqrt{k}}}.$$ 
Take to the power $k$, to obtain 
$$ (\frac{\sqrt{k}-1}{\sqrt{k}})^k\le e^{-\sqrt{k}}.$$ 
