# Summation of a Geometric Power Series

I was told that I could simplify the power series into a a^k geometric series by setting a= (6*e^t)/(11^n). I remember from Calculus 2 that I could calculate the sum of a geometric series in this form using the formula 1/1-r.

I don't remember how this is applicable to a situation in which I have multiple variables or how it would help me find the function as asked in the question.

The choice of $a$ you mention is not correct. The series is a function of $t$ and you are told this explicitly: "$M(t)$ is the power series." Therefore, the choice of $a$ must be such that it is independent of the index variable $n$; e.g., $$a = \frac{6 e^t}{11},$$ and then noting that $$\frac{5}{6} a^n = \frac{5}{6} \cdot \frac{6^n e^{nt}}{11^n} = \frac{5 \cdot 6^{n-1} e^{nt}}{11^n}.$$ Therefore, your series can be written $$M(t) = \frac{5}{6} \sum_{n=1}^\infty a^n = \frac{5}{6} \cdot \frac{a}{1-a}, \quad |a| < 1,$$ where $a$ is given above. Note that because the index variable begins at $n = 1$, the series has sum $a/(1-a)$, not $1/(1-a)$. The radius of convergence of this sum is $1$, hence the requirement $|a| < 1$, which leads to the criterion on $t$ for the original series to converge.
• As it is defined in the first equation, $a$ is a function of $t$; so then $M$ is also a function of $t$, and your solution should express $M$ in terms of $t$ rather than the intermediate/auxiliary variable $a$, which you introduced only for the convenience of calculation. – heropup Mar 17 '15 at 20:50
• Yes that is correct. Try the other parts of your question now that you have $M(t)$. – heropup Mar 17 '15 at 21:11
Hint: $\frac{5\cdot 6^{n-1}e^{nt}}{11^n}=\frac{5}{6}\cdot\left(\frac{6e^t}{11}\right)^n$