Consider the function defined sequentially by \[f_n(x)= \begin{cases} \dfrac{(-1)^n}{n} & \text{for }x\in\left(\dfrac{1}{n+1},\dfrac{1}{n}\right]\\[1ex] 0 & \text{else} \end{cases}\]where \(n\in\mathbb{N}\). What's the value of \[\int_0^1 f_n(x)\,dx~?\]

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Consider the function defined sequentially by \[f_n(x)= \begin{cases} \dfrac{(-1)^n}{n} & \text{for }x\in\left(\dfrac{1}{n+1},\dfrac{1}{n}\right]\\[1ex] 0 & \text{else} \end{cases}\]where \(n\in\mathbb{N}\). What's the value of \[\int_0^1 f_n(x)\,dx~?\]

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\dfrac{(-1)^n}{n^2(n+1)}\] ?
Yep, it reduces nicely into an infinite series. I'm curious to see how one might find the exact value of it.
^without prior knowledge of the polylogarithm. I've seen the answer on W|A.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

I thought the integral evaluates to that expression.. hmm let me go read it again :)
\[\int_0^1f_n(x)\,dx=\sum_{n=1}^\infty\left(\frac{(-1)^n}{n^2}+\frac{(-1)^{n+1}}{n(n+1)}\right)\](you can see this equivalence easily if you plot \(f_n\); rectangles of width \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1}{n(n+1)}\) and height \(\dfrac{(-1)^n}{n}\)) The second term gives a telescoping sum that's easy to deal with if you use the power series for \(-\ln(1+x)\). The first term is trickier.
1 Attachment
\[\int_0^1f_n(x)dx=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2(n+1)}\]\[=\sum_{n=1}^\infty(\frac{1}{(2n)^2}-\frac{1}{2n}+\frac{1}{2n+1})-\sum_{n=1}^\infty(\frac{1}{(2n-1)^2}-\frac{1}{2n-1}+\frac{1}{2n})\]\[=\frac{\pi^2}{24}+\ln(2)-1-(\frac{\pi^2}{8}-\ln(2))=\ln(4)-\frac{\pi^2}{12}-1\]Not sure if you can follow my thought process, but I THINK I got it haha
Well done! The thought to split into even and odd terms occurred to me a few moments ago. The rest is easy so long as we're familiar with the sum\[\sum_{n=1}^\infty \frac{1}{n^2}=\dfrac{\pi^2}{6}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question