Determine the series is absolutely convergent or diverges by using ratio test.
b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2} }$

= $\frac{e^2+e^ne^n+1}{e^2}$

is that the final answer? can it be simplified any further?

2. Originally Posted by anderson
Determine the series is absolutely convergent or diverges by using ratio test.
b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2} }$

= $\frac{e^2+e^ne^n+1}{e^2}$

is that the final answer? can it be simplified any further?

There are a number of errors, including:

1. The numerator of the second line of working should be a product not a sum.

2. $\lim_{n \to + \infty} \left(1+\frac{1}{n+1}\right)^{2n} \neq e^n e^n$.

3. Determine the series is absolutely convergent or diverges by using ratio test.
b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n^2+2n+1)}}{{(1+\frac{1}{n} )}^{2}}$
$\frac {{e}^{(n^2+2n+1)}}{{e}^{n^2}}$

i dont know how should i continue from here ..need some help & guidance for this question...for all i know the final answer is e, am wondering how to get the answer.

appreciate all your help & support.

4. Originally Posted by anderson
Determine the series is absolutely convergent or diverges by using ratio test.
b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\frac {{(1+\frac{1}{n+1})}^{n^2}+{(1+\frac{1}{n+1})}^{2n }+{(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2} }$

= $\frac{e^2+e^ne^n+1}{e^2}$

is that the final answer? can it be simplified any further?

$(1+\frac{1}{n})^{n^2}=\left((1+\frac{1}{n})^{n}\ri ght)^n$.
Apply the Root Test.

5. thank you for replying, not sure about the root test.can anyoine help me.
thank you for all your help & support.

i think i got the answer:
$

\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}}
$

${(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}$
=(e)(1)
=e

6. Originally Posted by anderson
thank you for replying, not sure about the root test.can anyoine help me.
thank you for all your help & support.

i think i got the answer:
$

\frac {{(1+\frac{1}{n+1})}^{n^2}{\color{green}{(1+\frac{ 1}{n+1})}^{2n}}{\color{red}{(1+\frac{1}{n+1})}^{2} }}{{(1+\frac{1}{n})}^{n^2}}
$

${(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}$
=(e)(1)
=e

Well.
the green one $\rightarrow e^2$
the red one $\rightarrow 1$ as $n\rightarrow\infty$.
so we can ignore the red one.
Now, we are left with $\lim_{n\to\infty} \frac{( 1 + \frac{1}{n+1})^{n^2} }{ (1 + \frac{1}{n})^{n^2}}$
Try to evaluate it, then multiply the result by $e^2$ to get the answer of the limit of the Ratio Test.

Actually, I did not know what are trying to use the Ratio Test.
It can be tested for convergence/divergence easily.
Clearly $\left( 1 + \frac{1}{n} \right)^{n^2} \rightarrow \infty$ as $n \rightarrow \infty$.
so the series diverges by the Test for Divergence (the nth term test).
Or note that:
$\left( 1 + \frac{1}{n} \right)^{n^2} \geq \left( 1 + \frac{1}{n} \right)^{n}$.
Then, the Basic Comparison Test will finish it.

7. hi General

this is the question:
Determine the series is absolutely convergent or diverges by using ratio test.
b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\lim_{n\to\infty}$ $\frac {{(1+\frac{1}{n+1})}^{n^2}{(1+\frac{1}{n+1})}^{2n} {(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}}$
= $\lim_{n\to\infty}$ ${(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}$
= $\lim_{n\to\infty}$ $(e)(1)$
=e

is it correct? can someone help to confirm?

8. Originally Posted by anderson
hi General

this is the question:
Determine the series is absolutely convergent or diverges by using ratio test.
b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\lim_{n\to\infty}$ $\frac {{(1+\frac{1}{n+1})}^{n^2}{(1+\frac{1}{n+1})}^{2n} {(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}}$
= $\lim_{n\to\infty}$ ${(1+\frac{1}{n+1})}^{2n}{(1+\frac{1}{n+1})}^{2}$
= $\lim_{n\to\infty}$ $(e)(1)$
=e

is it correct? can someone help to confirm?

Wrong.
You canceled two things. which is wrong. they are not equal.

9. hi General

thank you for replying.you are right. is this the correct working.

this is the question:
Determine the series is absolutely convergent or diverges by using ratio test.
b) $\sum ^\infty _{n=1}$ ${(1+\frac{1}{n})}^{n^2}$
$\frac {{(1+\frac{1}{n+1})}^{(n+1)^2}}{{(1+\frac{1}{n})}^ {n^2}}$
= $\lim_{n\to\infty}$ $\frac {{(1+\frac{1}{n+1})}^{n^2}{(1+\frac{1}{n+1})}^{2n} {(1+\frac{1}{n+1})}^{2}}{{(1+\frac{1}{n})}^{n^2}}$
= $\lim_{n\to\infty}$ $\frac{(e^2)(e)(1)}{e^2}$
=e
e>1, divergent.
thank you for all your help & guidance.really appreciate.

10. The limit = $e$, Right.
But sorry, am not sure if you really can cancel them or not.

See:
http://www.wolframalpha.com/input/?i...es+to+infinity

11. hi General

thank you for helping to confirm. i'm not sure about that either, but i dont know any other way to get the answer,e.

can anyone help to confirm this doubt? really appreciate all your help & contribution.