Question

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

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asked 2021-01-04
Using Maclaurin series, determine to exactly what value the following series converges:
\(\sum_{n=0}^\infty\frac{(\ln5)^n}{n!}\)

Answers (1)

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
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