# Math Help - Ratio test question

1. ## Ratio test question

hi everyone

need help with these questions.

Determine the series converges or diverges by using ratio test.

a) $\sum ^\infty _{n=1}$ $\frac {(-10)^n}{n!}$
= $\frac {\frac {(-10)^{(n+1)}}{(n+1)!}}{\frac {(-10)^n}{n!}}$
= $\frac{-10}{n+1}$
the series converges.

b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2} }$

i'm stuck on how to simplify,

really appreciate all your help & guidance.

thank you & regards.

2. Originally Posted by anderson
hi everyone

need help with these questions.

Determine the series converges or diverges by using ratio test.

a) $\sum ^\infty _{n=1}$ $\frac {(-10)^n}{n!}$
= $\frac {\frac {(-10)^{(n+1)}}{(n+1)!}}{\frac {(-10)^n}{n!}}$
= $\frac{-10}{n+1}$
the series converges.

b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2} }$

i'm stuck on how to simplify,

really appreciate all your help & guidance.

thank you & regards.
Try the root test for the second one. I'm sure you'll find a pleasant result.

3. A single little question... does the ratio test works on series of alternating terms?

4. Originally Posted by anderson
hi everyone

need help with these questions.

Determine the series converges or diverges by using ratio test.

a) $\sum ^\infty _{n=1}$ $\frac {(-10)^n}{n!}$
= $\frac {\frac {(-10)^{(n+1)}}{(n+1)!}}{\frac {(-10)^n}{n!}}$
= $\frac{-10}{n+1}$
the series converges.

b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2} }$

i'm stuck on how to simplify,

really appreciate all your help & guidance.

thank you & regards.
this "=" shoud not be there.
Its mean the sum of the series = the value of the limit of the ratio test!

5. Originally Posted by felper
A single little question... does the ratio test works on series of alternating terms?
Yes.
The absolute value in the limit of the ratio test will cancel the negative signs.
The same thing for the root test.

6. Originally Posted by General
Yes.
The absolute value in the limit of the ratio test will cancel the negative signs.
The same thing for the root test.
Aaahaa, i onle knew the positive terms version.