How do you solve for the sum of the series From the question$$\sum_{n=1}^{\infty}112*(0.065)^n$$My way of approaching is by trying to find the limit of the partial sum of a series but I can't get a general formula for the partial sum. Am I missing another easier way of approaching this?
Hint: The first term of the series is $112\times 0.065$ (not $112$). (Make sure you know why this is!) Then use the formula for an infinite geometric series.
According to the general expression for the partial sums of the geometric series:
$$\sum_{n=1}^N ar^n=ar\sum_{n=0}^{N-1} r^n=ar\frac {1-r^N}{1-r}.$$

If $|r|&lt;1$ (as in your example) the sum will tend to
$$\frac {ar}{1-r}. $$

Now substitute $a=112$, $b=0.065$.