Question

By recognizing each series below as a Taylor series evaluated at a particular value of x, find the sum of each convergent series.
A. 1 + 1/5 + (1/5)^2 + (1/5)^3 + (1/5)^4 +.....+ (1/5)^n + .... = _____.
B. 1 + 5 + 5^2/2! + 5^3/3! + 5^4/4! +....+ 5^n/n! +....= _____.

Answer 1

The first sum is a geometric series:

$1+\dfrac15+\dfrac1{5^2}+\dfrac1{5^3}+\cdots+\dfrac1{5^n}+\cdots=\displaystyle\sum_{n=0}^\infty\frac1{5^n}$

Recall that for |x| < 1, we have

$\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n$

Here we have |x| = |1/5| = 1/5 < 1, so the first sum converges to 1/(1 - 1/5) = 5/4.

The second sum is exponential:

$1+5+\dfrac{5^2}{2!}+\dfrac{5^3}{3!}+\cdots+\dfrac{5^n}{n!}+\cdots=\displaystyle\sum_{n=0}^\infty \frac{5^n}{n!}$

Recall that

$\exp(x)=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}$

which converges everywhere, so the second sum converges to exp(5) or e⁵.