The first sum is a geometric series:
[tex]1+\dfrac15+\dfrac1{5^2}+\dfrac1{5^3}+\cdots+\dfrac1{5^n}+\cdots=\displaystyle\sum_{n=0}^\infty\frac1{5^n}[/tex]
Recall that for |x| < 1, we have
[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]
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:
[tex]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!}[/tex]
Recall that
[tex]\exp(x)=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}[/tex]
which converges everywhere, so the second sum converges to exp(5) or e⁵.
