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Do these series converge or diverge using the root test

a) ∑n=1n1nn\displaystyle \sum_{n=1}^{n} \frac{1}{n^n}

lim\mathrm{lim}
Unparseable latex formula:

\limits_{n=\infty}

1nn1n\frac{1}{n^n}^\frac{1}{n} = lim\mathrm{lim}
Unparseable latex formula:

\limits_{n=\infty}

1n\frac{1}{n}
1n\frac{1}{n} is a harmonic series which diverges therefore the series ∑n=1n1nn\displaystyle \sum_{n=1}^{n} \frac{1}{n^n} diverges. Is this correct?

b) ∑n=1n2(n−1)nn\displaystyle \sum_{n=1}^{n} \frac{2^{(n-1)}}{n^n}
(edited 4 years ago)
so you need to examine the limit L of the nth root of the nth term ?

if L<1 then the series converges

if L>1 then the series diverges

if L = 1 we cannot tell if it converges or diverges
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Original post by the bear
so you need to examine the limit L of the nth root of the nth term ?

if L<1 then the series converges

if L>1 then the series diverges

if L = 1 we cannot tell if it converges or diverges


But i simplified (a) and it turns out to be a harmonic series so will the series diverge ?
Original post by E--
But i simplified (a) and it turns out to be a harmonic series so will the series diverge ?

certainly the sum of the harmonic series diverges....
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Original post by the bear
certainly the sum of the harmonic series diverges....

And therefore the sum of ∑n=1n1nn\displaystyle \sum_{n=1}^{n} \frac{1}{n^n} diverges?
Or is it that the limit of 1n\frac{1}{n} is 0 which is < 1 and therefore the series converges?
Original post by E--
And therefore the sum of ∑n=1n1nn\displaystyle \sum_{n=1}^{n} \frac{1}{n^n} diverges?
Or is it that the limit of 1n\frac{1}{n} is 0 which is < 1 and therefore the series converges?

the root test does not ask you to examine the sum of the roots of the nth terms, but the actual terms.
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Original post by the bear
the root test does not ask you to examine the sum of the roots of the nth terms, but the actual terms.

So you don't look at the sum of the series but the limit and in this case the limit is 0 so the series converges?
Original post by E--
So you don't look at the sum of the series but the limit and in this case the limit is 0 so the series converges?

Yes, the series converges.

I know the aim here is learning to use the root test, but this shouldn't stop you applying common sense.

It's obvious that (for n > 2) 1nn\dfrac{1}{n^n} is less than both 1n2\dfrac{1}{n^2} and 12n\dfrac{1}{2^n}, both of which are standard convergent series. So it must also converge by the comparison test.

Edit: to be clear, knowing it's smaller than *either* 1/n^2 or 1/2^n is sufficient to deduce convergence. I just wanted to point out there were *two* very obvious series to compare against here.
(edited 4 years ago)
Original post by E--
So you don't look at the sum of the series but the limit and in this case the limit is 0 so the series converges?


yes, that is right. Always take DFranklin's advice... he is a real expert :h:

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