# evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$

evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$

I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$

And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$

But how do I find the sum?

• Do you know geometric series? – k99731 Mar 2 '16 at 21:25
• You know the radius of convergence test but you forgot how to evaluate it. $\sum_{i=0}^n a^i = \frac{1-a^n}{1-a}$ and $\sum_{i=0}^{\infty} a^i = \frac 1{1-a}$ if $0<a<1$. – fleablood Mar 2 '16 at 23:27

Notice, $$\sum_{n=2}^{\infty}\frac{3}{10^n}$$ $$=3\sum_{n=2}^{\infty}\frac{1}{10^n}$$ $$=3\left(\underbrace{\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}+\ldots}_{\text{sum of an infinite G.P.}}\right)$$

$$=3\left(\frac{\frac{1}{10^2}}{1-\frac{1}{10}}\right)$$ $$=3\left(\frac{1}{90}\right)=\color{red}{\frac 1{30}}$$

The sum is $$0.03+0.003+0.0003+\ldots=0.03333\ldots=\frac1{30}$$

$$\sum^{\infty}_{n=2} \frac{3}{10^n}=\frac{3}{100}+\frac{3}{1000}+...=0.03+0.003+...=0.0333...=\frac{0.333...}{10}=\frac{\frac{1}{3}}{10}=\frac{1}{30}$$

You know that $\sum_{n\ge 1}q^n$ converges (you have $q=\frac{1}{10}$, similarily it works for all $q\in(-1,1)$), so let $$S=\sum_{n\ge 1}q^n=q+q^2+q^3+\dots$$ then $$qS=q^2+q^3+\dots = S-q$$ so$$S=\frac{q}{1-q}$$

Hence $$\sum_{n=2}^{\infty}\frac{3}{10^n}=3\left(\sum_{n=1}^{\infty}\left(\frac{1}{10}\right)^n-\frac{1}{10}\right)=\frac{1}{30}$$

Notice:

• Use the geometric series test to proof that this series converges;
• It converges when:

$$0<\left|\frac{1}{c}\right|<1$$

• $$\sum_{n=a}^{m}\frac{b}{c^n}=b\sum_{n=a}^{m}\frac{1}{c^n}=\frac{bc^{-a-m}\left(c^{1+m}-c^a\right)}{c-1}$$
• $$\sum_{n=a}^{\infty}\frac{b}{c^n}=\lim_{m\to\infty}\frac{bc^{-a-m}\left(c^{1+m}-c^a\right)}{c-1}=\frac{bc^{1-a}}{c-1}$$

So, solving your question, set $c=10$ to proof that this series converges:

$$\sum_{n=2}^{\infty}\frac{3}{10^n}=\lim_{m\to\infty}\frac{3\cdot10^{-2-m}\left(10^{1+m}-10^2\right)}{10-1}=\lim_{m\to\infty}\left[\frac{1}{30}-\frac{10^{-m}}{3}\right]=\frac{3\cdot10^{1-2}}{10-1}=\frac{1}{30}$$

Well $\sum_{i=1}^{\infty} 3/10^i = 0.333333.... = 1/3$.

So $\sum_{i=2}^{\infty} 3/10^i = \sum_{i=1}^{\infty} 3/10^i - 3/10 = 1/3 - 3/10= 1/30$

Hint: sum of a G.P. $$a+ar+ar^2+ar^3+\ldots+ar^{n}=\frac{a(r^{n+1}-1)}{r-1}$$