Question

How do you use the limit comparison test for `sum 1 / (n + sqrt(n))` for n=1 to `n=oo`?

Answer 1

`sum_(n=1)^oo1/(n+sqrt(n))` diverges, this can be seen by comparing it to `sum_(n=1)^oo1/(2n)`.

Explanation:

Since this series is a sum of positive numbers, we need to find either a convergent series `sum_(n=1)^(oo)a_n` such that `a_n>=1/(n+sqrt(n))` and conclude that our series is convergent, or we need to find a divergent series such that `a_n<=1/(n+sqrt(n))` and conclude our series to be divergent as well.

We remark the following:
For
`n>=1`, `sqrt(n)<=n`.
Therefore
`n+sqrt(n)<=2n`.
So
`1/(n+sqrt(n))>=1/(2n)`.

Since it is well known that `sum_(n=1)^oo1/n` diverges, so `sum_(n=1)^oo1/(2n)` diverges as well, since if it would converge, then `2sum_(n=1)^oo1/(2n)=sum_(n=1)^oo1/n` would converge as well, and this is not the case.

Now using the comparison test, we see that `sum_(n=1)^oo1/(n+sqrt(n))` diverges.

Answer 2

The limit comparison test takes two series, `suma_n` and `sumb_n` where `a_n>=0`, `b_ngt0`.

If `lim_(nrarroo)a_n/b_n=L` where `L>0` and is finite, then either both series converge or both series diverge.

We should let `a_n=1/(n+sqrtn)`, the sequence from the given series. A good `b_n` choice is the overpowering function that `a_n` approaches as `n` becomes large. So, let `b_n=1/n`.

Note that `sumb_n` diverges (it's the harmonic series).

So, we see that `lim_(nrarroo)a_n/b_n=lim_(nrarroo)(1/(n+sqrtn))/(1/n)=lim_(nrarroo)n/(n+sqrtn)`. Continuing by dividing through by `n/n`, this becomes `lim_(nrarroo)1/(1+1/sqrtn)=1/1=1`.

Since the limit is `1`, which is `>0` and defined, we see that `suma_n` and `sumb_n` will both diverge or converge. Since we already know at `sumb_n` diverges, we can conclude that `suma_n=sum_(n=1)^oo1/(n+sqrtn)` diverges as well.