Sum of Series 1/n: Is it Infinity?

[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+...

 

[jedishrfu]

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

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

 

[mfb]

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.

 

[Chris Miller]

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

So... it's < 2?

 

[mfb]

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

So... it's < 2?

It is.

 

[BWV]

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+...

the sum of a^k 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 a^k

 

[Matt Benesi]

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

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

 

[Chris Miller]

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

Same thing?

 

[Matt Benesi]

Same thing?

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.