Question

convergent divergent series

Answer 1

problems in file

Answer 2

@jim_thompson5910

Answer 3

\[11.\\
\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2}\]
Applying the ratio test, you have
\[\begin{align*}\lim_{n\to\infty}\frac{(2(n+1))!}{((n+1)!)^2}\cdot\frac{(n!)^2}{(2n)!} &= \lim_{n\to\infty} \frac{(2n+2)(2n+1)(2n)!(n!)^2}{((n+1)!)^2(2n)!}\\\\\\
&= \lim_{n\to\infty} \frac{(2n+2)(2n+1)n!n!}{(n+1)!(n+1)!}\\\\\\
&= \lim_{n\to\infty} \frac{(2n+2)(2n+1)(n!)^2}{(n+1)^2(n!)^2}\\\\\\
&= \lim_{n\to\infty} \frac{(2n+2)(2n+1)}{(n+1)^2}\\\\\\
&=4
\end{align*}\]
\(4>1\), which means the series diverges (by the ratio test).

Answer 4

And I'm assuming \(r\) is the limit, so that would mean \(r=4\) ?

Answer 5

yes

Answer 6

\[13.\\
\sum_{n=1}^\infty \frac{7}{[\ln (n+1)]^n}\]
Root test gives you
\[\lim_{n\to\infty}\sqrt[n]{\left|\frac{7}{[\ln(n+1)]^n}\right|}=\lim_{n\to\infty}\frac{7^{1/n}}{\ln(n+1)}=\frac{1}{\infty}=0\not=e\]

Answer 7

Even if you did get \(e\) as the limit, \(e>1\) would tell you that the series diverges, which it does not.

Answer 8

\[7.\\
a_n=\frac{n^{2.1}}{2.1^n}\]
\[\lim_{n\to\infty}\frac{(n+1)^{2.1}}{2.1^{n+1}}\cdot\frac{2.1^n}{n^{2.1}}=\frac{1}{2.1}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{2.1}=\frac{1}{2.1}\]
which means...

Answer 9

it converges I got 47619/100000

Answer 10

Right, you may have done a bit more work, you just have to show that it's smaller than 1.

Answer 11

what about number 5?

Answer 12

I haven't tried applying the ratio test to it, seems like too much of a hassle. But you're saying it converges, and I believe you (comparing with \(\sum\dfrac{1}{n^2}\)).

However, if a series \(\sum a_n\) converges, then you know that \(\displaystyle\lim_{n\to\infty}a_n=0\), so the last box shouldn't be checked.

Answer 13

thanks alot, you were really helpful.