- For the series \(\sum_{n=1}^{\infty} n x^n\), start from the geometric series \(\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\) and differentiate to obtain the sum \(\frac{x}{(1-x)^2}\).
- For the series \(\sum_{n=1}^{\infty} \frac{n}{5^n}\), evaluate the result of the first series at \(x = 1/5\).
- Simplify the expression: \(\frac{1/5}{(1-1/5)^2} = \frac{1/5}{16/25} = \frac{5}{16}\).
- The final answers are: (i) \(\boxed{\frac{x}{(1-x)^2}}\) and (ii) \(\boxed{\frac{5}{16}}\).
Explanation
- Sum of the series (i)
To find the sum of the series \(\sum_{n=1}^{\infty} n x^n\) for \(|x|<1\), we start with the power series expansion for a geometric series, which is known for \(|x| < 1\):
\[\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots = \frac{1}{1-x}\]
We differentiate both sides of this equation with respect to \(x\):
\[\frac{d}{dx} \left(\sum_{n=0}^{\infty} x^n\right) = \frac{d}{dx} \left(\frac{1}{1-x}\right)\]
\[\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}\]
To get the desired form, we multiply both sides by \(x\):
\[x \sum_{n=1}^{\infty} n x^{n-1} = x \frac{1}{(1-x)^2}\]
\[\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}\]
This confirms the sum of the series in (i).
- Sum of the series (ii)
For the series \(\sum_{n=1}^{\infty} \frac{n}{5^n}\), we notice that this is a specific case of the series we just analyzed, where \(x = \frac{1}{5}\).
Since the formula for the sum of the series is \(\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}\), we substitute \(x = \frac{1}{5}\) into this result:
\[\sum_{n=1}^{\infty} n \left(\frac{1}{5}\right)^n = \frac{\frac{1}{5}}{(1 - \frac{1}{5})^2}\]
Now, simplify the denominator:
\[1 - \frac{1}{5} = \frac{4}{5}\]
\[\left(\frac{4}{5}\right)^2 = \frac{16}{25}\]
Thus, the sum is:
\[\frac{\frac{1}{5}}{\frac{16}{25}} = \frac{1}{5} \times \frac{25}{16} = \frac{5}{16}\]
The sum of the series \(\sum_{n=1}^{\infty} \frac{n}{5^n}\) is \(\frac{5}{16}\).
- Final Conclusion
In conclusion, the sum of the series \(\sum_{n=1}^{\infty} n x^n\) for \(|x|<1\) is \(\frac{x}{(1-x)^2}\), and the sum of the series \(\sum_{n=1}^{\infty} \frac{n}{5^n}\) is \(\frac{5}{16}\).
Examples
Calculus series expansions are essential in various fields, such as probability and physics. For instance, determining the expected value of a geometric distribution, which represents the number of trials needed to get the first success, relies on summing series of the form \(\sum_{n=1}^{\infty} n p(1-p)^{n-1}\). Understanding these techniques allows scientists and statisticians to model systems efficiently and predict long-term behavior in repetitive processes.