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?
 A: 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. 
A: Observe that:
$$\sum_{n=1}^{\infty}a\cdot b^n=a\sum_{n=1}^\infty b^n$$
$$=a(b+b^2+...)=a(1+b+b^2+...)-a$$
Then use that $$x\neq 1 \implies 1+x+...+x^n=\frac{x^{n+1}-1}{x-1}$$
Hint: what is $\lim_{n\to\infty}b^n$, given some $0<b<1$?
A: From comments: "So I tried the infinite geometric series formula and got 119.7. I don't think that's right however"
$1 + 0.65 + 0.65^2 + ......  = \frac 1{1-0.065} \approx 1.070$ and so
$112( 1 + 0.65 + 0.65^2 + ......) \approx 119.7$
.... but...
also from comments.
"Be careful of the first term!"
$\sum_{k=0}^\infty 0.065^n= 1+0.065 + 0.065^2 + .... \approx 1.071$
but $\sum_{k=1}^\infty 0.065^n = 0.065 + 0.065^2 + 0.065^3 + ..... =$
either $(\sum_{k=0}^\infty 0.065^n) - 1$ or $0.065*(\sum_{k=0}^\infty 0.065^n)$.  Either way you get $\sum_{k=1}^\infty 0.065^n= \frac 1{1-0.065} -1 = \frac {0.065}{1-0.065} \approx 0.0695$ and 
Our answer is $\approx 7.786$.
A: $$\sum_{n=1}^{\infty}112 \cdot (0.065)^n = 112\sum_{n=1}^{\infty}0.065^n= 112 \cdot \frac {0.065}{1-0.065}$$
because $$\sum_{n=1}^{\infty}0.065^n$$ is a geometric series with the first element $a_1 = 0.065$ and the quotient $q_1 = 0.065$, too.
 Because $|q|<1$, the series is convergent, and we may use known formula
$$\sum = \frac {a_1}{1-q}$$
for it.
So the result is $$112\cdot \frac {65}{1000-65}=\frac {112\cdot65}{935}=\frac {112\cdot 13}{187}=\frac{1456}{187} \approx 7.78609625668449
 $$
A: 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 $|r|<1$ (as in your example) the sum will tend to
$$\frac {ar}{1-r}. $$
Now substitute $a=112$, $b=0.065$.
