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Advanced Knowledge Problem of the Week

Seems like it's yet another Wednesday, and that means it's time for yet another Advanced Knowledge Problem, which today is even more concise than this introduction.

Solved it or not, feel free to check out my solution, right below.

What did you think of this one? Did you solve it in a different way? Tell us about it in the comments section!

I did most of the same work; when I got to summation of 1/(n*(n-1)), I worked it from n=2 to constant i, determining it was (i-1)/i. I evaluated the limit as i -> infinity, using L'Hopital's Rule to get the answer of 1.

ReplyDeleteYou essentially break up the infinite sum

ReplyDelete\sum_{n=1}^\infty (1/n - 1/(n+1))

into the difference between two sums

\sum_{n=1}^\infty 1/n - sum_{n=1}^\infty 1/(n+1)

Each of these sums are divergent! Instead, you need to do something like

\sum_{n=1}^\infty (1/n - 1/(n+1))

= lim_{k\to\infty} \sum_{n=1}^k (1/n - 1/(n+1))

= lim_{k\to\infty} (1 - 1/(k+1))

= 1

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ReplyDeleteI like to learn that how to solve this equation.

ReplyDeleteI have break this infinite sum several times but didn't get the right answer.

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It is not as easy as you have made it seem. Appreciate the effort.

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