Series convergence proof and determining limit value

Question

I want to show that the following series is convergent and determine the corresponding limit value.

$$ a(n):=\sum_{k=1}^n \frac{k}{n^2}$$

To show the series converges I used the ratio test:

$$\left|\frac{k}{(n+1)^2}\right|\cdot\left|\frac{n}{k}\right|=\frac{n}{(n+1)^2}<1$$

So the series converges. But how can I go on now and calculate the limit value?

Hint: the $1/n^2$ does not depend on $k$ and can be factored out of the sum. — player3236, Dec 04 '20 at 15:42

score 1 · Accepted Answer · answered Dec 04 '20 at 15:53

1

$$\sum _{k=1}^n \frac{k}{n^2}=\frac{1}{n^2}\sum _{k=1}^nk=\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{n+1}{2n}\to\frac12$$

answered Dec 04 '20 at 15:53

Raffaele

1

Note that the question is a duplicate, a possible duplicate target had already been pointed out (and the same answer has been given before). – Martin R Dec 04 '20 at 15:55
@MartinR Sorry I didn't realize that – Raffaele Dec 04 '20 at 15:57
@MartinR I can't delete an accepted answer, tho – Raffaele Dec 04 '20 at 15:58

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