series and sequence

[naikn]

Determine wether the following series converges or diverges


Infinity
Sumation 2 + 3^n
4^n
n=1

I understand that you must split the sequence up into two parts so you will get 2 + (3/ 4)^n
4^n

I understand how to do the second part of the summation but am unable to figure out what to do with the first part

 

[daon]

\(\displaystyle 2\sum \frac{1}{4^n} + \sum (\frac{3}{4})^n\)

See it now?

Or notice that \(\displaystyle s_{n+1} > s_n\) and

\(\displaystyle n\ge 1 \implies \frac{2+3^n}{4^n} < \frac{3^n+3^n}{4^n} = 2(\frac{3}{4})^n\)

 

[BigGlenntheHeavy]

\(\displaystyle \sum_ {n=1}^{\infty} \frac{2+3^{n}}{4^{n}} \ = \ \sum_{n=1}^{\infty}\frac{2}{4^{n}} \ + \ \sum_{n=1}^{\infty}\frac{3^{n}}{4^{n}}\)

\(\displaystyle = \ 2 \sum_{n=1}^{\infty}(\frac{1}{4})^{n} \ + \ \sum_{n=1}^{\infty}(\frac{3}{4})^{n}\)

\(\displaystyle = \ 2\bigg[\frac{1/4}{1-1/4}\bigg] \ + \ \bigg[\frac{3/4}{1-3/4}\bigg] \ = \ \frac{11}{3}.\)

\(\displaystyle Hence, \ converges \ to \ \frac{11}{3}.\)