Convergence of series

Examine if the series:

$\displaystyle \sum_{n=1}^{\infty}\bigg ( e - \left ( 1+\tfrac{1}{n} \right )^n \bigg )$

converges.

Solution

We are using the basic inequality

$e^x \geq x +1 \quad \text{forall} \; x \in \mathbb{R}$

as well as the Hermite Hadamard inequality. Applying the Hermite – Hadamard inequality we have that

$\begin{align*} n \ln \left ( 1 + \frac{1}{n} \right ) &= n \int_{n}^{n+1} \frac{{\rm d}x}{x} \\ &\leq \frac{n}{2} \left ( \frac{1}{n} + \frac{1}{n+1} \right ) \\ &= 1 - \frac{1}{2n+2} \end{align*}$

Exponentiating we get that

$\begin{align*} \left ( 1 + \frac{1}{n} \right )^n & \leq e e^{-1/(2n+2)} \\ &\leq \frac{e}{1+ \frac{1}{2n+2}} \quad \quad (e^x \geq x +1)\\ &= e \left ( 1 -\frac{1}{2n+3} \right ) \end{align*}$

and hence

$\displaystyle e - \left ( 1 + \frac{1}{n} \right )^n \geq \frac{e}{2n+3}$

This leads us to the conclusion that the series diverges.

Convergence of series

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