Question

discuss the convergence of

Answer 1

\[1) \sum_{n=0}^{\infty} \frac{ 4^n }{ n! } \]

Answer 2

\[2) \sum_{n=2}^{\infty} \frac{ 1 }{ (\log n)^n }\]

Answer 3

@SithsAndGiggles

Answer 4

Use the ratio test for the first series:
\[\lim_{n\to\infty}\left|\frac{4^{n+1}}{(n+1)!}\cdot\frac{n!}{4^n}\right|\]
If the limit is less than 1, the series converges, and diverges otherwise.

Answer 5

oh thanks! n second?

Answer 6

I think the ratio test will also work for the second one.

Answer 7

\[\lim_{n\to\infty}\left|\frac{\left(\ln n\right)^n}{\left(\ln(n+1)\right)^{n+1}}\right|\]
Try it out, I haven't checked myself.

Answer 8

oh wow
i hv some bothersome sylabus i guess
i will try solving them! thanks!

Answer 9

@SithsAndGiggles answer to first was 0
2nd ..m still to calculate -.-

Answer 10

btw in second n=2, does that make any difference

Answer 11

No, the starting index won't affect the result of the test. The second series must start at \(n=2\) because \((\ln 0)^0\) and \((\ln1)^1=0\) would make the \(n=0\) and \(n=1\) terms undefined.

Answer 12

oh i see

Answer 13

@SithsAndGiggles i need hint for base of second,i dont remember any formula n google isnt helping

Answer 14

\[\lim_{n\to\infty}\left|\frac{\left(\ln n\right)^n}{\left(\ln(n+1)\right)^{n+1}}\right|=\lim_{n\to\infty}\frac{\left(\ln n\right)^n}{\left(\ln(n+1)\right)^{n+1}}\]
Hmm, I don't think there's an easy way to compute this, but you can reason that the denominator will always be larger than the numerator:
\[\begin{align*}\ln n&<\ln (n+1)\\
(\ln n)^n&<(\ln (n+1))^n\\
(\ln n)^n&<(\ln (n+1))^{n+1}
\end{align*}\]
So, since you have a ratio of the form \(\dfrac{a}{b}\) where \(b>a\), then as \(n\to\infty\) the limit must be 0.

Answer 15

oh thats alot easier
thanks