# How to prove the given series is divergent?

Given series $$\sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right],$$ try to show that it is divergent!

The criterion will show that it is the case of limits $1$, so maybe need some other methods? any suggestions?

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Hint: Use the Comparison Test, the series diverges. – Amzoti Dec 11 '12 at 15:46

Using the binomial theorem, we get that for $n\ge1$, \begin{align} \left(1+\frac1n\right)^{\large n} &=\sum_{k=0}^n\frac{\binom{n}{k}}{n^k}\\ &=\sum_{k=0}^\infty\frac1{k!}\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\cdots\frac{n-k+1}{n}\\ &\le2+\left(1-\frac1n\right)\sum_{k=2}^\infty\frac1{k!}\\ &=2+\left(1-\frac1n\right)(e-2)\\ &=e-\frac{e-2}{n} \end{align} Therefore, for $n\ge1$, $$e-\left(1+\frac1n\right)^{\large n}\ge\frac{e-2}{n}$$ So by comparison with the Harmonic series, $$\sum_{n=1}^\infty\left(e-\left(1+\frac1n\right)^{\large n}\right)$$ diverges.

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Let $x > 1$. Then the inequality $$\frac{1}{t} \leq \frac{1}{x}(1-t) + 1$$ holds for all $t \in [1,x]$ (the right hand side is a straight line between $(1,1)$ and $(x, \tfrac{1}{x})$ in $t$) and in particular $$\log(x) = \int_1^x \frac{dt}{t} \leq \frac{1}{2} \left(x - \frac{1}{x} \right)$$ for all $x > 1$. Substitute $x \leftarrow 1 + \tfrac{1}{n}$ to get $$\log \left(1 + \frac{1}{n} \right) \leq \frac{1}{2n} + \frac{1}{2(n+1)}$$ and after multiplying by $n$ $$\log\left(1 + \frac{1}{n} \right)^n \leq 1 - \frac{1}{2(n+1)}.$$ Use this together with the estimate $e^x \leq (1-x)^{-1}$ for all $x < 1$ to get $$\left(1 + \frac{1}{n} \right)^n \leq e \cdot e^{-\displaystyle\frac{1}{2(n+1)}} \leq e \cdot \left(1 - \frac{1}{2n+3} \right)$$ or $$e - \left(1 + \frac{1}{n} \right)^n \geq \frac{e}{2n+3}.$$ This shows that your series diverges.

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That's great! The first inequality is more technical, how do you figure it out? – van abel Dec 17 '12 at 7:46

Note that \begin{align} \sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right]\geq & \sum_{n=1}^{\infty}\left[\sum_{k=0}^{n}\frac{1}{k!}-\left(1+\frac{1}{n}\right)^{n}\right] & \quad \forall n\in\mathbb{N} \\ = & \sum_{n=1}^{\infty}\left[\sum_{k=0}^{n}\frac{1}{k!}- \sum_{k=1}^{n}\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right] & \forall n\in\mathbb{N} \\ = & \sum_{n=1}^{\infty}\sum_{k=0}^{n}\left[\frac{1}{k!}-\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right] & \forall n\in\mathbb{N} \\ \geq & \sum_{n=1}^{\infty} \left[\frac{1}{k!}-\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right] & \\ \geq & \sum_{n=1}^{\infty} \left[\frac{1}{k!}\left(\frac{1}{n}\right)^k-\frac{n!}{(n-k)!k!}\left(\frac{1}{n}\right)^k\right] & \\ \geq & \sum_{n=1}^{\infty} \left[\frac{1}{k!}-\frac{n!}{(n-k)!k!}\right]\cdot\left(\frac{1}{n}\right)^k & \end{align} for all $k\in\{0,1,\dots,n\}$. If $k=1$, \begin{align} \sum_{n=1}^{+\infty}\left[e-\left(1+\frac{1}{n}\right)^{n}\right]\geq & \sum_{n=1}^{\infty} \left|1-n\right|\cdot\left(\frac{1}{n}\right) & \geq\sum_{n=1}^{\infty}\frac{1}{2}=\infty \end{align}

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just the last step is $1/n-1>-1/2$, which on the other hand cann't show that it is divgence. – van abel Dec 11 '12 at 16:45
@vanabel Thank's. – MathOverview Dec 11 '12 at 17:09