Question

# Using Maclaurin series, determine to exactly what value the following series converges: sum_{n=0}^inftyfrac{(ln5)^n}{n!}

Series
Using Maclaurin series, determine to exactly what value the following series converges:
$$\sum_{n=0}^\infty\frac{(\ln5)^n}{n!}$$

2021-01-05
Maclaurin Series expansion of $$e^x$$
Let $$f(x)=e^x$$
$$f(x)=f(0)+f'(0)\frac{x}{1!}+f''(x)\frac{x^2}{2!}+f'''(x)\frac{x^3}{3!}+...$$
$$f(0)=e^0=1$$
$$f'(0)=e^0=1$$
$$f''(0)=e^0=1$$
$$f'''(0)=e^0=1$$... ans so on
Hence,
$$e^x=1+\frac{x}{1}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
Now we have given series $$\sum_{n=0}^\infty\frac{(\ln5)^n}{n!}$$
If we compare it equation, we got here $$x=\ln5$$
$$\Rightarrow\sum_{n=0}^\infty\frac{(\ln5)^n}{n!}=e^{\ln5}$$
$$=5$$
Hence $$\sum_{n=0}^\infty\frac{(\ln5)^n}{n!}=5$$
So , this series converges to 5