Root Test — Question 9

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Question 9

Determine whether ∑n=1∞n−n\displaystyle\sum_{n=1}^{\infty}n^{-n} converges or diverges. Use the Root Test and explain why the exponent nn makes this test especially efficient.

Original worksheet page 1: question and worked solution for 4-11-009
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Question 9 – Solution

Step 1: Rewrite the term.

We have an=n−n=(1/n)na_n=n^{-n}=(1/n)^n.

Step 2: Take the nnth root.

|an|n=(1n)nn=1n.\sqrt[n]{|a_n|}=\sqrt[n]{\left(\frac1n\right)^n}=\frac1n. Therefore L=limn→∞1n=0<1.L=\lim_{n\to\infty}\frac1n=0<1.

Conclusion.

The series converges absolutely. Rather than complicating the calculation, the exponent nn is exactly canceled by the nnth root. Moreover, for any 0<r<10<r<1, eventually 1/n<r1/n<r, so n−n<rnn^{-n}<r^n and the tail is dominated by a geometric series.

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