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

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