Question

How do you use the limit comparison test to determine if `Sigma n/((n+1)2^(n-1))` from `[1,oo)` is convergent or divergent?

Answer 1

`sum_(n=1)^oon/((n+1)2^(n-1))`

Rewriting this:

`=sum_(n=1)^oon/(n+1)1/(2^(n-1))=sum_(n=1)^oon/(n+1)1/(2^n/2)=sum_(n=1)^oon/(n+1)2/2^n`

Bringing the `2` out and noticing that `1/2^n=(1/2)^n`:

`=2sum_(n=1)^oon/(n+1)(1/2)^n`

We should recognize that `sum_(n=1)^oo(1/2)^n` is a geometric series, and since `1/2<1`, we know this series will converge.

Also note that when `n>0`, all terms of `n/(n+1)<1`. This means we can say that:

`n/(n+1)(1/2)^n<=(1/2)^n`

We can now use the direct comparison test. Since `sum_(n=1)^oo(1/2)^n` converges, and `n/(n+1)(1/2)^n<=(1/2)^n`, we know that `sum_(n=1)^oon/(n+1)(1/2)^n` converges as well.

Thus `2sum_(n=1)^oon/(n+1)(1/2)^n=sum_(n=1)^oon/((n+1)2^(n-1))` is convergent.