Does this sequence converge or diverge?

Converges to 0.

A sequence converges if it has a restrict. If it does not have a limit, it diverges. There isn't any single experiment that is going to instantly inform you what happens for any given sequence. You must take matters on a case by means of case basis. On this case, one test we are able to use the evaluation experiment. If we can find a converging sequence the place each time period is better than the corresponding term of our sequence, then our sequence have got to converge too. I do know (1 + (1/n) )^n converges to e (2.71828...), in view that that's the definition of e. On the grounds that 1/n > 1/2n > 0 for n >=1, this means 1+1/n > 1 + 2/n > 1, so (1 + 1/n)^n > (1 + 1/2n)^n. So the sequence converges. On a facet observe, it doesn't converge to 1, as someone else prompt. It without a doubt converges to sqrt(e).

Use the fact that

1 + 2 + 3 + ... + n

equals

n * (n+1) / 2.

Then you can write

a_n = (n * (n+1) / 2) / n^2

a_n = 1/2 + 1/(2n)

As n -> infinity, 1/(2n) approaches 0. Therefore, the sequence converges and the limit is

1/2 + 0 = 1/2.

This sequence is equivalent to

a(n) = (1+2+3+4+...+n)/(n^2)

Now, 1+2+3+4+...+n = (n)(n+1)/2 = (n^2+n)/2

So the sequence is equivalent to:

(n^2 + n)/(2n^2)

Which equals

n^2/(2n^2) + n/(2n^2)

Which equals

1/2 + 1/(2n)

As n -> infinity this converges to 1/2 (because 1/(2n) -> 0)

a_n = (1^1)/(n^2) + (2^1)/(n^2) + ... + (n^1)/(n^2)

= (1 + 2 + ... + n)/(n^2)

= [n(n+1)/2]/(n^2)

= (n+1)/(2n)

= 1/2 + 1/(2n).

Therefore, a_n converges to 1/2.

Lord bless you today!

Your sequence is 1/n and converges to zero.

===============

(n^1)/n² reduces to 1/n (Now see "examples" in link below.)

The SERIES (sum of the terms in the sequence) diverges.