Kind of like a special product, except it's a sum

Algebra Level 3

$\large \displaystyle\sum_{n=1}^{2015}\left ( \frac{n}{n+1} \right )^{(-1)^{n}} = \ ?$

The answer is 2016.

2 solutions

Maggie Miller
Jul 29, 2015

For even $n$ , $\left(\frac{n}{n+1}\right)^{(-1)^n}+\left(\frac{n+1}{n+2}\right)^{(-1)^{n+1}}=\frac{n}{n+1}+\frac{n+2}{n+1}=2.$

Therefore, $\displaystyle\sum_{n=1}^{2015}\left(\frac{n}{n+1}\right)^{(-1)^n}=\frac{2}{1}+\left(\frac{2014}{2}\right)\cdot2=\boxed{2016}$ .

Moderator note:

Yes. Taking advantage of the simple fact that $(-1)^n = -(-1)^{n+1}$ is very useful.

Yes! Incidentally, the 'special product' referred to in the title is the Wallis Product . The addends in this problem are the same as the factors of the Wallis Product. The interesting thing about the Wallis Product is that it converges to $\frac{\pi}{2}$ .

Andy Hayes - 5 years, 10 months ago

Cool, I didn't know about that!

Maggie Miller - 5 years, 10 months ago

Dammit! You stole my punchline :(

(The guy in my picture is John Wallis)

It's a really cool question though Andy. Keep them coming!

Isaac Buckley - 5 years, 10 months ago

Used the same method.

Niranjan Khanderia - 5 years, 10 months ago

Aakash Khandelwal
Jul 29, 2015

if you know the next year then you may guess the answer

Anyone has the chance to guess an answer and get it right. The primary challenge of this problem is to interpret the notation. The pattern is not that difficult to find once you write out the first couple of terms and group the terms with like denominators. Also note that the series only evaluates to a whole number when the final $n$ is odd.

Andy Hayes - 5 years, 10 months ago

Kind of like a special product, except it's a sum

The answer is 2016.

2 solutions

Moderator note:

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