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I am trying to calculate the sum of this infinite series: $$\sum\limits_{n=1}^{\infty \:\:}\frac{2n-1}{5^{2n-1}}$$

From ratio test, I knew that the series is convergent. But I still can't figure out how to find its sum. Is there any technique to tackle this?

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  • $\begingroup$ evaluate these series $$a:= \sum_{n\geqslant 1}\frac{n}{5^n}, \quad b:=\sum_{n\geqslant 1}\frac{2n}{5^{2n}}$$ then your series is $a-b$. There are other more direct ways also. $\endgroup$
    – Masacroso
    Commented May 28, 2020 at 15:57

3 Answers 3

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After a small rewriting: $$ \frac{2n-1}{5^{2n-1}}=\frac{2n -1}{25^n\div 5}\\ =10\frac{n}{25^n}-5\frac{1}{25^n} $$ These two terms can be summed separately using known formulas. We have $$ \sum_{n=1}^\infty\frac{n}{25^n}=\frac{25}{576}\\ \sum_{n=1}^\infty\frac1{25^n}=\frac1{24} $$

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Split the series into two:

$$\sum_{n=1}^{\infty} \frac{2n}{5^{2n-1}}+\sum_{n=1}^{\infty} \frac{-1}{5^{2n-1}}$$

The first series is in the form: $$S\left(1-\frac{1}{25}\right)=\frac{\frac{2}{5}}{1-\frac{1}{25}}$$ Note: This can be derived by writing the first few terms of the series then subtracting that initial series, $S$, by $S\cdot\frac{1}{25}$ in order to create a geometric series where the numerator does not change.

The second series is in the form: $$S=\frac{-\frac{1}{5}}{1-\frac{1}{25}}$$.

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Hint:

\begin{align} \sum_{n=1}^\infty (2n-1)x^{2n-1}&=x\sum_{n=1}^\infty (2n-1)x^{2n-2}=x\biggl(\sum_{n=1}^\infty x^{2n-1}\biggr)'\\ &=x\biggl(x\sum_{n=1}^\infty x^{2n-2}\biggr)'=x\biggl(\frac x{1-x^2}\biggr)' =\frac{x(1+x^2)}{(1-x^2)^2}. \end{align}

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