Science:Math Exam Resources/Courses/MATH101/April 2015/Question 05 (b)

MATH101 April 2015

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Question 05 (b)
The series $\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{(2n+1)^{2}}}$ converges to some number S (you don’t have to prove this). According to the Alternating Series Estimation Theorem, what is the smallest value of n for which the nth partial sum of the series is at most ${\frac {1}{100}}$ away from S? For this value of n, write out the nth partial sum of the series.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall the Alternating Series Estimation Theorem: suppose an alternating series $\sum _{n=1}^{\infty }(-1)^{n+1}a_{n}$ converges to $S$. Let $S_{n}$ be the nth partial sum of the series. Then, if the sequence $\{a_{n}\}$ is positive for any $n\geq 1$, monotonically decreasing and converges to 0, then we have $|S-S_{n}|\leq a_{n+1}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since $a_{n}={\frac {1}{(2n+1)^{2}}}$ is positive for any $n\geq 1$, monotonically decreasing, and converges to 0, applying the Alternating Series Estimation Theorem, we have $|S-S_{n}|\leq a_{n+1}={\frac {1}{(2(n+1)+1)^{2}}}={\frac {1}{(2n+3)^{2}}}.$ Now, it is enough to find the smallest $n$ such that ${\frac {1}{(2n+3)^{2}}}\leq {\frac {1}{100}}.$ By taking square roots on both side, the inequality is equivalent with ${\frac {1}{2n+3}}\leq {\frac {1}{10}}$ and hence with $10\leq 2n+3$. Therefore, we can easily obtain the smallest natural number $n$ as 4 satisfying $10\leq 2n+3$. The answer: $n=\color {blue}4$ and the partial sum is $S_{4}=\color {blue}{\frac {1}{3^{2}}}-{\frac {1}{5^{2}}}+{\frac {1}{7^{2}}}-{\frac {1}{9^{2}}}$