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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.

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