Question

series n=0 to infinity ((e^n)+3)/(4^n-1) does it converge and what is its sum?

Answer 1

\[\int\limits_{0}^{\infty}\frac{ e ^{n}+3 }{ 4^{n}-1 }\]

Answer 2

is that correct?

Answer 3

@Kanwar245 Summation,

Answer 4

oh right

Answer 5

no the it is 4^(n-1) the 1 is in the exponent

Answer 6

\[\sum_{n=0}^{\infty}\frac{ e ^{n}+3 }{ 4^{n-1} }\] ?

Answer 7

yeah

Answer 8

ok so each term is positive

Answer 9

use ratio or root test

Answer 10

use root test

Answer 11

\[\lim_{n \rightarrow \infty} \sqrt[n]{a _{n}}\]

Answer 12

\[\lim_{n \rightarrow \infty}\sqrt[n]{\frac{ e ^{n}+3 }{ 4^{n-1} }}\]

Answer 13

i separated the the fraction into two fractions and used geometric rule, but having trouble with it.

Answer 14

okay so when you separate it, on the first part use root test, and the second part is a geometric series of 1/4 which converges

Answer 15

first part would also converge by root test

Answer 16

I don't know how would you find the sum of the series though

Answer 17

for that you would need to look at the partial sums and find their sums

Answer 18

\[a _{n}=3^{n+1}/2^{n}\]