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?

how do i simplifly this question,please help. really appreaciate all your help & support

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?

how do i simplifly this question,please help. really appreaciate all your help & support
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?

how do i simplifly this question,please help. really appreaciate all your help & support
$(1+\frac{1}{n})^{n^2}=\left((1+\frac{1}{n})^{n}\ri ght)^n$.
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.
Sorry, I did not follow your solution. show work please.
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

thank you for replying.

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?

thank you in advance for all your help & guidance.

8. Originally Posted by anderson
hi General

thank you for replying.

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?

thank you in advance for all your help & guidance.
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.