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CheemnCatelvew

Answered question

2021-01-04

Using Maclaurin series, determine to exactly what value the following series converges:

\sum_{n = 0}^{\infty} \frac{(\ln 5)^{n}}{n!}

Obiajulu

Skilled2021-01-05Added 98 answers

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{(\ln 5)^{n}}{n!}

If we compare it equation, we got here

x = \ln 5

\Rightarrow \sum_{n = 0}^{\infty} \frac{(\ln 5)^{n}}{n!} = e^{\ln 5}

= 5

Hence

\sum_{n = 0}^{\infty} \frac{(\ln 5)^{n}}{n!} = 5

So , this series converges to 5

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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