Question

Find the smallest value of \(n\) so that \(R_n\le10^{-6}\).

\(\sum_{k=1}^{\infty}\frac{k}{e^k}\)

Short answer

The smallest value of \(n\) so that \(R_n\le10^{-6}\) holds is \(n=17\).

Step-by-step solution

Step 1. Given Information

The series is \(\sum_{k=1}^{\infty}\frac{k}{e^k}\) .

Step 2. Check for intergral test

Consider function \(f\left(x\right)=xe^{-x}\) .

\(f'\left(x\right)=e^{-x}-xe^{-x}\)

\(=\left(1-x\right)e^{-x}\)

The derivative is negative for for all \(x>1\) .Therefore, the function is decreasing.

The function \(f\left(x\right)=xe^{-x}\) is continuous, decreasing, with positive terms.

Therefore, all the conditions of integral test are fulfilled. So, integral test is applicable.

Step 3. Determine Convergence or divergence

Consider the integral \(\int_{x=1}^{\infty}f(x)dx=\int_{x=1}^{\infty}xe^{-x}dx \).

To evaluate the integral, integrate by parts.

Therefore,

\(\int_{x=1}^{\infty}xe^-x dx =\lim_{k\to\infty} \int_{x=1}^{k}xe^-x dx \)

\(=2e^{-1}\)

\(=\frac{2}{e}\)

The integral converges. Therefore, the series \(\sum_{k=1}^{\infty}\frac{k}{e^k}\) is convergent.

Step 4. Find value of \(n\)

The objective is to use the \(10^{th}\) term in the sequence of partial sums to approximate the sum of the series.

The strategy to find the \(10^{th}\) term in the sequence of partial sum is to expand the series for \(k\) from \(1\) to \(10\).

To compute the \(10^{th}\) term \(S_{10}\), use the series \(\sum_{k=1}^{\infty}\frac{k}{e^k}\).

The value of \(S_{10}\) is:

\(S_{10}=\frac{1}{e}+\frac{2}{e^2} +\frac{3}{e^3}+\ldots+\frac{10}{e^{10}}\)

\(= 0.3679 + 0.2708 + 0.1494 + 0.0733 + 0.0337 + 0.0149+0.0064 + 0.0027 + 0.0011 + 0.0004\)

\(\approx0.9206\)

Therefore, the \(10^{th}\) term in the sequence of partial sums to approximate the sum of the series is \(S_{10}\approx0.9206\).

If a function \(a\) is continuous, positive, and decreasing, and if the improper integral \(\int_{1}^{\infty}a\left(x\right)\) converges, then the \(n^{th}\) remainder, \(R_{n}\) for the series \(\sum_{k=1}^{\infty}a\left(k\right)\) is bounded by:

\(0\le R_{n}\le\int_{n}^{\infty}a\left(x\right)dx\)

Furthermore, if \(B_{n}=\int_{n}^{\infty}a\left(x\right)dx\) , then

\(S_{n}\le\sum_{k=1}^{\infty}a\left(k\right)\le S_{n}+B_{n}\)

Therefore, \(0\le R_{10}\le\int_{x=10}^{\infty}xe^{-x}dx\)

The value of integral \(\int_{x=10}^{\infty}xe^{-x}dx\) is:

\(\int_{x=10}^{\infty}xe^{-x}dx=10e^{-10}+e^{-10}\)

\(=0.0004\)

Therefore, \(0\le R_{10}\le0.0004\)

Assume \(B_{10}\approx0.1\) and \(\sum_{k=1}^{\infty}a\left(k\right)\).

Using the enquality \(S_{n}\le\sum_{k=1}^{\infty}a\left(k\right)\le S_{n}+B_{n}\) , the following inequality is obtained.

\(S_{10}\le L\le S_{10}+B_{10}\)

\(0.9206\le L\le0.9206 + 0.0004\)

\(0.9206\le L\le0.9206 + 0.921\)

Therefore, the interval containing the sum of the series is \(L\in\left(0.9206,0.921\right)\)

To find the smallest value of \(n\) find \(n\) such that following holds:

\(\int_{n}^{\infty}a\left(x\right)dx\le10^{-6}\)

\(ne^{-n}+e^{-n}\le10^{-6}\)

\(\left(n+1\right)e^{-n}\le10^{-6}\)

\(n\ge16.6884\)

Therefore, the value of \(n\) so that \(R_{n}\le10^-6\) holds is \(n=17\).