Question

Is the following series convergent? If so find its sum.
Im posting the series now.....

Answer 1

\[\sum_{n=1}^{\infty} (1 + 7^{n}) / 9^{n}\]

Answer 2

I find the limit of the sequence to be zero which I guess means its convergent? But I dont know how to find the sum.

Answer 3

the sum is found by pretty much separating this into a few parts and using summation rules

Answer 4

\[\sum_{n=1}^{\infty} \frac{1 + 7^{n}}{9^{n}}\]
\[\sum_{n=1}^{\infty} \frac{1}{9^{n}} + \frac{7^{n}}{9^{n}}\]
\[\sum_{n=1}^{\infty} \frac{1}{9^{n}} + \sum_{n=1}^{\infty}\frac{7^{n}}{9^n}\]
\[\sum_{n=1}^{\infty} (\frac{1}{9})^n + \sum_{n=1}^{\infty}(\frac{7}{9})^n\]

Answer 5

the sum of a geometric series is defined as:\[a\frac{1-r^n}{1-r}\]
where a is starting constant, and r is the common ration

these have no starting constants that need to be addressed so we can simply go with:

\[\frac{1-(\frac{1}{9})^n}{1-\frac{1}{9}}+\frac{1-(\frac{7}{9})^n}{1-\frac{7}{9}}\]

and the nth terms we already determined go to zero leaving us with the rest of it to calculate

Answer 6

from your last summation line can I say that \[\sum_{n=1}^{\infty} (1/9)^{n} = \sum_{n=1}^{\infty} (1/9)(1/9)^{n-1}\]

Answer 7

you can if you like; i do notice that there is a slight difference between starting this at n=0 and n=1 that needs an adjustment

Answer 8

the adjustment amounts to there being a starting "a"; 1/9 for the left and 7/9 for the right

Answer 9

\[a_1\frac{1-(\frac{1}{9})^n}{1-\frac{1}{9}}+a_2\frac{1-(\frac{7}{9})^n}{1-\frac{7}{9}}\]
\[\frac{1}{9}\frac{1-(\frac{1}{9})^n}{1-\frac{1}{9}}+\frac{7}{9}\frac{1-(\frac{7}{9})^n}{1-\frac{7}{9}}\]

Answer 10

better :)

Answer 11

thanks