limit of series exponential 
Compute the limit of the series $$\sum\limits_{n=4}^\infty 3\frac{2^{n+1}}{5^{n-2}}$$
How do you approach these types of problems?
I'm thinking that this one is in indeterminate form, is that correct?
 A: HINT:
So, the $r\ge 4$ the term  $$t_r=3\cdot \frac{2^{r+1}}{5^{r-2}}=3\cdot \frac 2{5^{-2}}\cdot \left(\frac25\right)^r=150\cdot \left(\frac25\right)^r$$
Clearly, this is an  infinite geometric series with the common ratio is $\frac25$ and the first term $=150\cdot \left(\frac25\right)^4$
A: The biggest hint lies on the fact that you have a $2^n$ in the numerator and $5^n$ on the denominator.  
So if you are able to manipulate the series so that it kind of looks like $$\sum_{n=0}^\infty ({2\over5})^n$$, then you can use the fact that this is an infinite geometric sequence with the common ratio having an absolute value less than 1. 
Thus you can find the infinite sum to be $$1 \over {1 - {2\over 5}}$$
Now let's move on to how to manipulate it. First we want to get rid of the numbers around $({2 \over 5})^n$, so it will go as follows. $$3 {{2^{n+1}}\over{5^{n-2}}}
 = 3 {2 \over 5^{-2}}・{{2^n}\over{5^n}}$$ $$ = 150 ・ ({2 \over5})^n$$
so the original series can be expressed as $$150\sum_{n=4}^{\infty}({2\over 5})^n $$
Now what we need to do is to deal with the index of the sum, which oddly starts from 4. The following is what we usually do $$\sum_{n=4}^{\infty}({2\over 5})^n $$ is basically the same thing as the geometric sum, but it's "missing" the first four terms, as in $$\sum_{n=0}^{\infty}({2\over 5})^n = 1 + {2 \over 5} + ({2\over 5})^2 + ({2\over 5})^3 + \sum_{n=4}^{\infty}({2\over 5})^n$$
So calling the first four terms $S$ for convenience, the following is what you will need.
$$\sum_{n=4}^{\infty}3 {{2^{n+1}}\over{5^{n-2}}} = 150 \left[ \sum_{n=0}^{\infty}({2\over 5})^n - S \right]$$
Most problems that seems like it's related to geometric series goes like this.
