Euler And Series

Yahoo Answers is a goldmine. There are a lot of great questions in there. Today I have one relating to series and Euler. Yes, this guy again. He always comes back for more.

A sequence a_n is defined as 1/n²(n + 1)². Find the sum of the series from 1 to infinity.

Remember Euler from before? He proved that the sum of 1/n² from 1 to infinity is π²/6. Yes, Euler was a smart man. All we need now is to transofrm a_n to something with 1/n² which includes sums. But how?

Let's start with the obvious and proceed from there. To create that denominator, we can try to add 1/n² and 1/(n + 1)², see what we get, and then decide what else to do. So:
1/n² + 1/(n + 1)² = (n + 1)²/n²(n + 1)² + n²/n²(n + 1)²

Which is:
(n² + (n + 1)²)/n²(n + 1)²

Open parentheses and add:
(n² + n² + 2n + 1)/n²(n + 1)²
(2n² + 2n + 1)/n²(n + 1)²

To leave only the 1 in the numerator (which will give a_n, we need to subtract (2n² + 2n)/n²(n + 1)²:
a_n = (2n² + 2n + 1)/n²(n + 1)² - (2n² + 2n)/n²(n + 1)²

The second term can be also written as 2n(n + 1), which can be cancelled. At the end, we get:
a_n = (2n² + 2n + 1)/n²(n + 1)² - 2/n(n + 1)

2/n(n + 1) can also be written as (2n + 2 - 2n)/n(n + 1), which can be transformed into:
(2n + 2)/n(n + 1) - 2n/n(n + 1) = 2/n - 2/(n + 1)

Since it's the opposite on a_n, we can finally conclude that:
a_n = 1/n² + 1/(n + 1)² + 2/(n + 1) - 2/n

Now it's really easy. Notice how 2/(n + 1) and -2/n cancel each other, except for the first term. Their sum is -2/1 + 2/2 - 2/2 + 2/3 - 2/3 + 2/4 - 2/4 + ... . Since 2/∞ is zero, the only remainder of this part of the series is -2.

As Euler said, the sum of the 1/n² part of the series is π²/6. The 1/(n + 1)² has the exact same sum, but minus one. Since the sums starts from 1/2² and not 1/1², the 1/1² is left outside.

The total sum of a_n from 1 to infinity is:
π²/6 + π²/6 - 1 - 2
S = π²/3 - 3

Since the terms of a_n get very small very quickly, it can be easily approximated for validity.

Hope you like it,
Nadav

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