series and sequence

naikn

New member
Joined
Aug 17, 2009
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1
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
 
214n+(34)n\displaystyle 2\sum \frac{1}{4^n} + \sum (\frac{3}{4})^n

See it now?

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

n1    2+3n4n<3n+3n4n=2(34)n\displaystyle n\ge 1 \implies \frac{2+3^n}{4^n} < \frac{3^n+3^n}{4^n} = 2(\frac{3}{4})^n
 
n=12+3n4n = n=124n + n=13n4n\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}}

= 2n=1(14)n + n=1(34)n\displaystyle = \ 2 \sum_{n=1}^{\infty}(\frac{1}{4})^{n} \ + \ \sum_{n=1}^{\infty}(\frac{3}{4})^{n}

= 2[1/411/4] + [3/413/4] = 113.\displaystyle = \ 2\bigg[\frac{1/4}{1-1/4}\bigg] \ + \ \bigg[\frac{3/4}{1-3/4}\bigg] \ = \ \frac{11}{3}.

Hence, converges to 113.\displaystyle Hence, \ converges \ to \ \frac{11}{3}.
 
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