# Behavior of :$\sum_{n=k}^{+\infty} n^{-\frac{n}{\log n}}$ with $k$ is natural number

I'm confused about behavior of this series :$\sum_{n=k}^{+\infty} n^{-\frac{n}{\log n}}$ with $k$ is natural number , for $k=0 ,1$ the series converges according to wolfram alpha assumption however $k=0 ,1$ are singular points , Now I want to know more about behavior of that series playing with values of the integer $k$ , for $k=1$ the series converges and it gives:

$$\sum_{n=1}^{+\infty} n^{-\frac{n}{\log n}}=\frac{1}{e-1}\tag{1}$$ and for $k=2$ we have : $$\sum_{n=2}^{+\infty} n^{-\frac{n}{\log n}}=\frac{1}{(e-1)e}\tag{2}$$ and for $k=3$ we have : $$\sum_{n=3}^{+\infty} n^{-\frac{n}{\log n}}=\frac{1}{(e-1)e^2}\tag{3}$$ , I have conjuctered that : $$\sum_{n=k}^{+\infty} n^{-\frac{n}{\log n}}=\frac{1}{(e-1)e^{k-1}}\tag{4}$$ , Now I have tow question :

Question:01 : Why wolfram alpha assumed that series converges for $k=0,1$ however they are singular points ?

Question:02

Does what i have conjuctered in $4$ true ?

## 2 Answers

Hint. The series is geometric: for $n> 1$, $$n^{-\frac{n}{\log n}}=\exp\left(-\frac{n}{\log n}\cdot \log(n)\right)=(1/e)^n.$$

• What about n=0,1 ? – zeraoulia rafik Jun 30 '18 at 18:59
• For $n=0$ and $n=1$ the term $n^{-\frac{n}{\log n}}$ is not defined. – Robert Z Jun 30 '18 at 19:19
• But wolfram alpha assumed that convergent – zeraoulia rafik Jun 30 '18 at 19:43
• @zeraouliarafik The singularities at $0$ and $1$ are removable and once removed, the series are defined as WA reports. – Mark Viola Jun 30 '18 at 19:48
• Yes, they are removable, but formally $n^{-\frac{n}{\log n}}$ is not defined at $0$ and at $1$. – Robert Z Jun 30 '18 at 20:24

HINT:

$$n^{-n/\log(n)}=e^{-n}$$

Now sum the geometric series