# B summation of series

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1. Feb 13, 2019 at 9:00 AM

### Chris Miller

I've seen the proof that the sum of 1/n for = 1 to infinity is infinity (which still blows my mind a little).

Is the sum of 1/nn for n = 1 to infinity also infinity?

i.e, 1 + 2/4 + 3/27 + 4/256+...

2. Feb 13, 2019 at 9:03 AM

### Staff: Mentor

I don't think you mean $n^n$ where $n=1$ right?

also your series should read $1 + 1/4 + 1/27 ...$?

3. Feb 13, 2019 at 9:04 AM

### Staff: Mentor

After the first two elements it is smaller than the sum over $\displaystyle \frac{1}{2^n}$ where you can find the sum directly.

If you want to explore the region between divergence and convergence more, have a look at $\displaystyle \frac{1}{n^2}$ and $\displaystyle \frac{1}{n \log(n)}$ and $\displaystyle \frac{1}{n \log(n)^2}$and $\displaystyle \frac{1}{n \log(n) \log(log(n))}$. These will need some clever tricks to determine if they converge.

4. Feb 13, 2019 at 9:19 AM

### Chris Miller

oops, yes, thanks, I meant the sum of n/nn for n = 1 to infinity

So... it's < 2?

5. Feb 13, 2019 at 9:29 AM

### Staff: Mentor

$\frac{n}{n^n}$? That is equal to $\frac{1}{n^{n-1}}$
It is.

6. Feb 13, 2019 at 1:27 PM

### BWV

the sum of ak converges for a<1 for k=1 to inf - this is just a geometric series and the sum is 1/(1-a)

so by that 1/nn converges for n=1 to inf as 1/nn becomes smaller than any constant ak

7. Feb 13, 2019 at 2:09 PM

### Matt Benesi

He meant $\sum\limits_{n\to\infty} \,\, \frac{n}{n^n}$.

8. Feb 13, 2019 at 2:57 PM

### Chris Miller

Same thing?

9. Feb 14, 2019 at 1:22 PM

### Matt Benesi

Unless you guys are mind readers, and meant $\sum\limits_{n\to\infty} \,\, \frac{n}{n^n} = \sum\limits_{n\to\infty} \,\, \frac{1}{n^{n-1}}$ by what you said, $\frac{n}{n^n}\, =\, \frac{1}{n^{n-1}}$ is different from the sum.

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