Questions: ∑(k=1 to ∞) k^2 / e^k

Solution Steps

To approximate the sum of the infinite series \(\sum_{k=1}^{\infty} \frac{k^{2}}{e^{k}}\), we can compute the partial sum up to a large number of terms. This approach leverages the fact that the terms decrease rapidly due to the exponential in the denominator.

The given series is

\[ \sum_{k=1}^{\infty} \frac{k^{2}}{e^{k}} \]

This is an infinite series where each term is of the form \(\frac{k^2}{e^k}\).

This series is a type of exponential series. The general form of an exponential series is

\[ \sum_{k=1}^{\infty} \frac{k^n}{a^k} \]

where \(a = e\) and \(n = 2\) in this case.

The series

\[ \sum_{k=1}^{\infty} \frac{k^n}{a^k} \]

can be evaluated using the formula for the exponential generating function:

\[ \sum_{k=1}^{\infty} \frac{k^n}{a^k} = \frac{Li_n(1/a)}{a} \]

where \(Li_n(x)\) is the polylogarithm function. For \(n = 2\), this becomes:

\[ \sum_{k=1}^{\infty} \frac{k^2}{e^k} = \frac{Li_2(1/e)}{e} \]

The polylogarithm function \(Li_2(x)\) is a special function that can be evaluated numerically. For \(x = 1/e\), it is known that:

\[ Li_2(1/e) \approx 0.822467 \]

Substitute the value of \(Li_2(1/e)\) into the series formula:

\[ \sum_{k=1}^{\infty} \frac{k^2}{e^k} = \frac{0.822467}{e} \]

Since \(e \approx 2.7183\), we have:

\[ \frac{0.822467}{2.7183} \approx 0.3020 \]

Final Answer