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I can't figure out whether this series is convergent.

I'm trying to use d'Alembert or Cauchy ratio tests, but however far I go with Taylor series it always ends up being one. The series is :

$$\sum_{n=1}^\infty \big((1+1/n)^n-e\big)$$

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  • $\begingroup$ use binomial expansion $\endgroup$ – mathworker21 Sep 22 at 23:18
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    $\begingroup$ If memory serves me correcty, the sequence $(1+1/n)^n - e$ doesn't approach $0$ very quickly. I believe it's in $O(1/n)$ which would make the series comparable to the harmonic series. $\endgroup$ – Robert Wolfe Sep 22 at 23:41
  • $\begingroup$ @RobertWolfe $(1+1/n)^n -e= e^{n\ln\left(1+\frac{1}{n}\right)}-e = e^{n\left(\frac{1}{n}-\frac{1}{2n^2}+\mathcal{O}(n^{-3})\right)}-e=e\left(e^{-\frac{1}{2n}+\mathcal{O}(n^{-2})}-1\right)$ so $\sum_{n=1}^\infty \big((1+1/n)^n-e\big)$ converges iff $\sum_{n=1}^\infty \left(e^{-\frac{1}{2n}}-1\right)$ converges but $e^{-1/n} -1 \sim -\tfrac{1}{2k} +\mathcal{O}(n^{-2})$ so the sum diverges as the harmonic series. $\endgroup$ – Brevan Ellefsen Sep 23 at 1:22
  • $\begingroup$ @RobertWolfe Alternatively, we can just explicitly work out the Laurent Series implicitly derived above, namely $(1+1/n)^n \sim e - \tfrac{e}{2n} + \mathcal{O}(n^{-2})$, via basic complex analysis and we get the same result $\endgroup$ – Brevan Ellefsen Sep 23 at 1:24
  • $\begingroup$ The result actually diverges to $-\infty$, yo may see my answer posted u below $\endgroup$ – Dr Zafar Ahmed DSc Sep 23 at 3:45
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As an answer hasn't been accepted yet, I assume that there's something left to be desired. So I'll take a different stab at the problem that only involves calculus.

We first observe, as in the comments, that $$\Bigl(1+\frac{1}{n}\Bigr)^n-e=e^{n\ln(1+1/n)}-e\,.$$

Elementary calculus arguments can show that the function $x\ln(1+1/x)$ is monotone increasing on $(0,\infty)$ and that its range has $1$ as a supremum and has $0$ as an infimum.

We now invoke the Mean Value Theorem to conclude that for every $n\in\mathbb{N}$ there is a real $\xi_n$ between $1$ and $n\ln(1+1/n)$ such that $$e-\Bigl(1+\frac{1}{n}\Bigr)^n=e^{\xi_n}\bigl(1-n\ln(1+1/n)\bigr)\,.$$

From what we know about $x\ln(1+1/x)$, we can conclude that $$e^0\bigl(1-n\ln(1+1/n)\bigr)\leq e-\Bigl(1+\frac{1}{n}\Bigr)^n\leq e^1\bigl(1-n\ln(1+1/n)\bigr)\,.$$

Now we use the inequalities \begin{equation}\frac{1}{2(1+x)}\leq 1-x\ln(1+1/x)\leq\frac{1}{x+1}\end{equation} which is true for all positive $x$ to conclude $$\frac{1}{2(1+n)}\leq e-\Bigl(1+\frac{1}{n}\Bigr)^n\leq\frac{e}{1+n}$$ which will carry the rest of the argument.

The real crux to this argument are the inequalities \begin{equation}\frac{1}{2(1+x)}\leq 1-x\ln(1+1/x)\leq\frac{1}{x+1}\,.\end{equation}

The right-hand inequality is equivalent to the more familiar inequality $$\ln(1+x)\leq x$$ which I can derive if needed.

The left-hand inequality is equivalent to the slightly stronger inequality $$\ln(1+x)\leq x-\frac{x^2}{2(1+x)}$$ which is however established in very much the same way as the weaker inequality.

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    $\begingroup$ Ninad's answer actually shows that $e-(1+1/n)^n\sim e/(2n)$. Although I haven't proved it, it seems that $$\frac{e}{2(x+1)}\leq e-(1+1/x)^x\leq\frac{e}{2x+1}$$ for all positive $x$, which gives very little wiggle room. $\endgroup$ – Robert Wolfe Sep 24 at 18:16
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Take the negative of that series to make the terms positive and limit compare it with $\frac{1}{n}$

$$\lim_{n\to\infty} \frac{e-\left(1+\frac{1}{n}\right)^n}{\frac{1}{n}} \to \frac{0}{0}$$

$$\implies \lim_{n\to\infty} \frac{-\left(1+\frac{1}{n}\right)^n\Biggr(\log\left(1+\frac{1}{n}\right)-\frac{1}{n+1}\Biggr)}{-\frac{1}{n^2}}$$

which we get from L'Hopital and logarithmic differentiation. If this limit exists, we can decompose it into a product. The limit on the left is $e$, so let's evaluate

$$\lim_{n\to\infty} \frac{\log\left(1+\frac{1}{n}\right)-\frac{1}{n+1}}{\frac{1}{n^2}}\to \frac{0}{0}$$

$$\implies \lim_{n\to\infty} \frac{\frac{1}{1+\frac{1}{n}}\cdot\left(-\frac{1}{n^2}\right)+\frac{1}{(n+1)^2}}{-\frac{2}{n^3}} = \lim_{n\to\infty}\frac{1}{2}\frac{n^2}{(n+1)^2} = \frac{1}{2}$$

This means the original limit equals $\frac{e}{2}$, so the series behaves as $\frac{1}{n}$, i.e. it diverges by the limit comparison test.

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  • $\begingroup$ Damn, you were faster than me 😅 Also, it should say over $\frac1n$ in the first row $\endgroup$ – Maximilian Janisch Sep 22 at 23:44
  • $\begingroup$ @MaximilianJanisch haha next time it'll be the other way around I was typing this up for 30 minutes as you can see by the typo I was trying a few other approaches, too $\endgroup$ – Ninad Munshi Sep 22 at 23:47
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Let $u=1/n$ then the Mc Laurin expansion is $$(1+u)^{1/u}-e-eu/2+11eu^2/24+...$$

Then $$\sum_{n=1}^{\infty} [\left(1+\frac{1}{n}\right)^n-e]=\sum_{n=1}^{\infty} \left( e-\frac{e}{2n}+\frac{11 }{24 n^2}-...-\frac{e}{2} \right)=-\frac{e}{2}\sum_{n=1}^{\infty} \frac{1}{n}+ \frac{11e}{24}\sum_{n=1}^{\infty}\frac{1}{n^2}+... $$ $$\sim -\frac{e}{2} \sum_{1}^ {\infty} \frac{1}{n} = -\infty$$ Here the very first series being divergent the given sum diverges to $-\infty$.

We may also write that $$\sum_{n=1}^{N}\left[\left (1+\frac{1}{n} \right)^n-e \right] \sim \frac{-e}{2} H_N \sim \frac{-e}{2} \ln N.$$ where $H_n$ is the sum of the Harmonic series. Further, you may see the plot of $$S(N)=\sum_{n=1}^{N}\left[\left (1+\frac{1}{n} \right)^n-e \right]$$ vs $N$ below Plot of <span class=$S(N)$">

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