B Summation of series

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+...
 
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I don't think you mean ##n^n## where ##n=1## right?

also your series should read ##1 + 1/4 + 1/27 ... ##?
 
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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.
 
oops, yes, thanks, I meant the sum of n/nn for n = 1 to infinity

So... it's < 2?
 

BWV

375
279
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 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
 

Matt Benesi

Gold Member
133
7
##\frac{n}{n^n}##? That is equal to ##\frac{1}{n^{n-1}}##It is.
He meant [itex]\sum\limits_{n\to\infty} \,\, \frac{n}{n^n}[/itex].
 

Matt Benesi

Gold Member
133
7
Same thing?
Unless you guys are mind readers, and meant [itex]\sum\limits_{n\to\infty} \,\, \frac{n}{n^n} = \sum\limits_{n\to\infty} \,\, \frac{1}{n^{n-1}}[/itex] by what you said, [itex]\frac{n}{n^n}\, =\, \frac{1}{n^{n-1}}[/itex] is different from the sum.
 

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