Question

How do you test for convergence for ?

`sum sqrt(n + 1)/(n^2+1)` for n=1 to infinity?

Answer 1

You have to think that when `n` is large `sqrt(n+1)` 'behaves' like `sqrt(n)` and `n^2+1` 'behaves like `n^2`

Explanation:

When `n` is large `sqrt(n+1)` grows 'at the same rate' than `sqrt(n)`. Formally, the `lim_(n->oo) sqrt(n+1)/sqrt(n)=1`

Similarly, `n^2+1` grows 'the same' as `n^2` because

`lim_(n->oo) (n^2+1)/n^2=1`

(This is often expressed as `sqrt(n+1)~~sqrt(n)` and `n^2+1~~n^2` when `n->oo`)

Now, we can see (informally) that the given series behaves like

`sum(sqrt(n))/n^2=sum 1/n^(3/2)` which is convergent because the exponent at the denominator is `>1`. So we estimate that our series is convergent.

Now we have to prove that this is true. Because we estimate that the series is convergent, we are going to compare it with a larger series that is convergent.

But since `n>=1` we can say:

`sqrt(n+1)<=sqrt (n+n)=sqrt(2n)`,

and since always `n^2+1 >n^2` we can write:

`sum sqrt(n+1)/(n^2+1)<=sum sqrt(2n)/n^2=sum sqrt(2)*sqrt(n)/(n^2)=sum sqrt(2)*1/n^(3/2)=sqrt(2) sum 1/n^(3/2)`
which is convergent, so our original series is also convergent.