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

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 :

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

• use binomial expansion – mathworker21 Sep 22 at 23:18
• 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. – Robert Wolfe Sep 22 at 23:41
• @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. – Brevan Ellefsen Sep 23 at 1:22
• @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 – Brevan Ellefsen Sep 23 at 1:24
• The result actually diverges to $-\infty$, yo may see my answer posted u below – Dr Zafar Ahmed DSc Sep 23 at 3:45

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.

• 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. – Robert Wolfe Sep 24 at 18:16

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.

• Damn, you were faster than me 😅 Also, it should say over $\frac1n$ in the first row – Maximilian Janisch Sep 22 at 23:44
• @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 – Ninad Munshi Sep 22 at 23:47

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 $S(N)$">