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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.

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$\ln(1+x)\le x$ 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 $\sum_{k=1}^{+\infty}\left(\frac{\sqrt k-1}{\sqrt k}\right)^k<+\infty$.

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  • $\begingroup$ Where did your k^2 term come from? $\endgroup$ – Walt Aug 6 at 16:46
  • $\begingroup$ I use that if $k^2a_k\to0$ when $k\to+\infty$, then $\sum_{k}a_k<+\infty$. $\endgroup$ – Will Aug 6 at 16:51
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Hint $(1-\frac{1}{\sqrt{k}})^{k}=((1-\frac{1}{\sqrt{k}})^{\sqrt{k}})^{\sqrt{k}}\rightarrow e^{-\sqrt{k}}$

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    $\begingroup$ This seems like poor reasoning. Using the same type of reasoning I could say that $(1-\frac{1}{n})^n\rightarrow 1^n$ which is certainly false. $\endgroup$ – Walt Aug 6 at 16:47
  • $\begingroup$ A sloppy answer. C'mon... $\endgroup$ – Wlod AA Aug 7 at 4:21
  • $\begingroup$ The result is correct. $\endgroup$ – marty cohen Aug 8 at 23:53
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$\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.

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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:

$$\begin{equation} \begin{aligned} \text{Your Sum} < \sum_{k=1}^\infty \exp(-\sqrt{k}) &< \sum_{k=0}^\infty \int \limits_{k}^{k+1} \exp(-\sqrt{r}) dr \\[6pt] &= \int \limits_0^\infty \exp(-\sqrt{r}) dr \\[6pt] &= \Bigg[ - 2(1+\sqrt{r}) \exp(-\sqrt{r}) \Bigg]_{r=0}^{r \rightarrow \infty} \\[6pt] &= \Bigg[ 0 - ( - 2 ) \Bigg] = 2 < \infty. \\[6pt] \end{aligned} \end{equation}$$

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  • $\begingroup$ T is decreasing and $e^{-\sqrt{k}} > e^{-k}$. $\endgroup$ – marty cohen Aug 8 at 23:49
  • $\begingroup$ @martycohen: Thanks for pointing our the error. I have re-worked the proof to correct. $\endgroup$ – Reinstate Monica Aug 9 at 1:37
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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

$$ (\frac{\sqrt{k}-1}{\sqrt{k}})^k\le e^{-\sqrt{k}}.$$

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