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Chapter 9: Problem 69

Identify the two series that are the same. (a) \(\sum_{n=1}^{\infty} \frac{n 5^{n}}{n !}\) (b) \(\sum_{n=0}^{\infty} \frac{n 5^{n}}{(n+1) !}\) (c) \(\sum_{n=0}^{\infty} \frac{(n+1) 5^{n+1}}{(n+1) !}\)

Short Answer

Expert verified
Series (a), (b), and (c) are all identical after suitable adjustments to their indices and limits.

Step by step solution

01

Evaluate series (a)

Given the series (a): \(\sum_{n=1}^{\infty} \frac{n 5^{n}}{n !}\). If we observe the series, we can see the factor of \(n\) multiplying with the power of \(5^n\) and dividing with \(n!\). Based on this, it becomes evident that we do not need to alter series (a) and can therefore always keep it as it is.
02

Evaluate series (b)

Evaluate the series (b): \(\sum_{n=0}^{\infty} \frac{n 5^{n}}{(n+1) !}\). This series begins at \(n = 0\), but \(n\) multiplies the terms in the numerator while in the denominator we have \((n+1)!\). We have to adjust it to make the factorial consistent. First step in analyzing: at \(n = 0\), the term becomes 0, so we can change the limit to \(n = 1\). Then replace \(n\) by \(n-1\) to match the format of the factorial in the denominator: \(\sum_{n=1}^{\infty} \frac{(n-1) 5^{n-1}}{n !}\). This series, after alteration, is identical with series (a).
03

Evaluate series (c)

Observe series (c): \(\sum_{n=0}^{\infty} \frac{(n+1) 5^{n+1}}{(n+1) !}\). Start by examining: at \(n = 0\), the term becomes \(5\), To align this series with series (a), replace \(n+1\) by \(n\) as this will correctly match the factorial in the denominator and adjust the power of 5: \(\sum_{n=1}^{\infty} \frac{n 5^{n}}{n !}\). This series, after alteration, is equivalent to series (a) and series (b).

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Most popular questions from this chapter

Proof In Exercises \(13-16,\) prove that the Maclaurin series for the function converges to the function for all \(x .\) $$ f(x)=\cos x $$

Find the values of \(x\) for which the series converges. $$\sum_{n=0}^{\infty}\left(\frac{x-3}{5}\right)^{n}$$

Verifying a Formula In Exercises 45 and \(46,\) use a power series and the fact that \(i^{2}=-1\) to verify the formula. $$ g(x)=\frac{1}{2 i}\left(e^{i x}-e^{-i x}\right)=\sin x $$

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