Question

Examine the convergence of series

Answer 1

a) \[\sum_{n=1}^{+\infty} \frac{ (-1)^n }{ 2^n }(\frac{ n+1 }{ n })^( n ^{2})\]

Answer 2

b)
\[\sum_{n=1}^{+\infty} \frac{ (-1)^n n^3 }{ 3^n }(x+3) ^{n}\]

Answer 3

what do you need to do?

Answer 4

i need to examine the convergence

Answer 5

Have you tried the ratio test?

Answer 6

well for the first one i have n+1 and i know that 1/n+1 = \[\sum_{n=0}^{\infty}(-1) ^n x^n\]

Answer 7

i can do the absolute of everything and then i get
\[\sum_{n=1}^{\infty}(\frac{ n+1 }{ n })^2*\frac{ 1 }{ 2^n }\]

Answer 8

\(|\frac{ (-1)^{n+1} (n+1)^3 (x+3) ^{n+1}}{ 3^{n+1} }* \frac{ 3^n}{(-1)^n n^3 (x+3) ^{n} }|=\\|\frac{-1(x+3)(n+1)^3}{3n^3}|=|\frac{-1(x+3)}{3}(\frac{n+1}{n})^3|\rightarrow |\frac{-x}{3}-1|\)

Note that the only way this will converge

Answer 9

is when \(|\frac{-x}{3}-1|<1\)

Do something similar for the other one.

Answer 10

how did i get -x/3 -1 ?

Answer 11

can i use cauchy for the 1st one so i get rid of n^2
and then ratio and then i just divide everything with n^2?

Answer 12

\(\lim_{n\rightarrow \infty }|\frac{-1(x+3)}{3}(\frac{n+1}{n})^3|=\lim_{n\rightarrow \infty }(\frac{n+1}{n})^3|\frac{-1(x+3)}{3}|=\\1*|\frac{-x-3}{3}|=|\frac{-x}{3}-1|\)

Answer 13

I have not worked the first one out, but show your work and if it works... it works :)

Answer 14

must sleep, good luck.

Answer 15

could you just check out the first one, i will post it now

Answer 16

\[c=\lim \sqrt[n^2]{(\frac{ n+1 }{ n })^2} \frac{ 1 }{ 2^n } = \lim \frac{ n+1 }{ n }\frac{ 1 }{ 2^n }\]

Answer 17

after that \[\lim \frac{ n+2 }{( n+1 )*2^n }*\frac{ n2^n }{ n+1 }=\frac{ n(n+2) }{ 2(n+1)^2 }\]

Answer 18

and after i put n-> intfty, i get 0 which is less than 1

Answer 19

still need me to look?