Solved: "Does ∑n=1∞1n1+1n converge? If yes, to where?"

Shamar Padilla

Answered question

2022-01-22

Does

\sum_{n = 1}^{\infty} \frac{1}{n^{1 + \frac{1}{n}}}

converge? If yes, to where?

saennwegoyk

Beginner2022-01-23Added 7 answers

The series does not converge.
Consider whether

n^{\frac{1}{n}} \geq k

for some constant k. If

k = e

, we have

\frac{1}{n} \ln n \geq 1 \to \ln n \geq n

But for all

n \geq 1, \ln n < n

, so we have an upper bound on

n^{\frac{1}{n}}

which means that the original sum is asymptotic to

\sum_{n = 1}^{\infty} \frac{1}{n}

mihady54

Beginner2022-01-24Added 13 answers

You can try the limit comparison theorem (more general form):

\frac{\frac{1}{n \sqrt[n]{n}}}{\frac{1}{n \log n}} = \frac{\log n}{\sqrt[n]{n}} \to \infty

and then your series diverges since the other diverges.
For the other one: you can use Cauchys

RizerMix

Expert2022-01-27Added 656 answers

2^{n} \geq n

for

n \geq 1

is easy to show. Hence

2 \geq n^{1 / n}

, so

\frac{1}{n^{1 + 1 / n}} \geq \frac{1}{2 n}

which implies

\sum \frac{1}{n^{1 + 1 / n}} \geq \frac{1}{2} \sum \frac{1}{n}

, which diverges.

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