I was to depict the convergence & divergence nature of the summation $\sum A_n$ where $A_n = (n^{1/n}-1)^k$ I was able to prove that when $k>1$ then $\sum A_n$ is converging and while $k<0$ it was diverging but when $1>k>0$ I saw the sequence $A_n$ to be increasing but I am unable to prove if the summation $\sum A_n$ is converging or diverging, so how do I prove it ?

  • $\begingroup$ You always have $A_n > 0$ for $k>0$. Thus, if the sequence is increasing, it can't be a null sequence. What does that tell you? $\endgroup$ – PhoemueX Sep 2 '15 at 19:56
  • $\begingroup$ True if it's increasing then it must be divergent but then I only observed it whether by cauchy's ratio test or D'alembert's root test I am unable to prove if it's divergent or not $\endgroup$ – Arnav Das Sep 2 '15 at 20:00

Since $e^x\geq x+1$,

$$ n^{\frac{1}{n}}-1 = \exp\left(\frac{\log n}{n}\right)-1 \geq \frac{\log n}{n} $$ hence if $0<\alpha\leq 1$, $$ \sum_{n=1}^{N}\left(n^{1/n}-1\right)^\alpha \geq \sum_{n=1}^{N}\frac{\log(n)^\alpha}{n^\alpha} $$ and the last series is diverging as $N\to +\infty$ by comparison with the generalized harmonic series.


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