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Determine whether ∑ 1 / n1 + 1/n converges

Test the following series for convergence:

    \[ \sum_{n=1}^{\infty} \frac{1}{n^{1 + \frac{1}{n}}}. \]

The series diverges.

Proof. First, we write

    \[ \frac{1}{n^{1+\frac{1}{n}}} = \frac{1}{n \cdot n^{\frac{1}{n}}}. \]

Then, we know n < 2^n for all n \geq 1. (We can deduce this from the Bernoulli inequality, (1+x)^n \geq 1 + nx with x = 1. We proved the Bernoulli inequality in this exercise, Section I.4.10, Exercise #14.) Therefore, n^{\frac{1}{n}} < 2 and we have

    \[ \frac{1}{n \cdot n^{\frac{1}{n}}} > \frac{1}{2n}. \]

Since the series \sum \frac{1}{2n} diverges we have established the divergence of the given series

    \[ \sum_{n=1}^{\infty} \frac{1}{n^{1+\frac{1}{n}}}. \qquad \blacksquare\]

One comment

  1. Evangelos says:

    I like the throwback to the Introductory section. Personally, I’d have used a simple limit comparison test to show.

        \begin{align*} \frac{1}{n^{1 + 1/n}} \sim \frac{1}{n}\ as \ n \rightarrow \infty \end{align*}

    But I’d still say yours wins out for effort.

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