Need Help #Series

I was going through some proofathon problems and came over this: ∑i=0∞∑j=0∞1(2i)!(2j+1)!\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{1}{(2i)!(2j+1)!} . I know we have to use the exponential series in here. But I am still skeptical about my answer. I am getting my answer : (e2+1)(e2−1)4e2\frac{(e^{2}+1)(e^{2}-1)} {4e^{2}} . (Which I think is probably wrong. :P) Would be glad if someone would explain me how to approach such kind of problems. (I always stuck at double summation problems. :P)

#Calculus #DoubleSum #InfiniteSeries

Note by Sanchit Ahuja
6 years, 3 months ago

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Comments

Actually, double summation always doesn't involves applying some formula or some difficult series expansion. The real technique for solving most of the double summations lies in the fact that you need to recognize the series expansion. For this I would recommend to first have a good skills (an practice) of simple sums. Indeed learn to recognize series summations by practising more and more simple summation.

For example - the above sum can be evaluated easily if you remember that ex=∑n=0∞xnn!e^x = \sum_{n=0}^\infty \frac{x^n}{n!}

Now in the given sum, both ii and jj are independent variables, so we can write the given sum as

∑i=0∞∑j=0∞1(2i)!(2j+1)!=(∑i=0∞1(2i)!)(∑j=0∞1(2j+1)!)\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{1}{(2i)!(2j+1)!} = \left(\sum_{i=0}^\infty \frac{1}{(2i)!}\right) \left(\sum_{j=0}^\infty \frac{1}{(2j+1)!}\right)

Now, the two sums are respectively the sum of reciprocal of even and odd factorials. TO find them we use the series expansion ee and e−1e^{-1}

e=∑n=0∞1n!e = \sum_{n=0}^\infty \frac{1}{n!}

e−1=∑n=0∞(−1)nn!e^{-1} = \sum_{n=0}^\infty \frac{(-1)^n}{n!}

Adding both of them gives

e+e−1=2∑n=0∞1(2n)!e+e^{-1} = 2\sum_{n=0}^\infty \frac{1}{(2n)!}

Similarly subtracting the second one from first one gives

e−e−1=2∑n=0∞1(2n+1)!e-e^{-1} = 2\sum_{n=0}^\infty \frac{1}{(2n+1)!}

Thus, our required sum is ∑i=0∞∑j=0∞1(2i)!(2j+1)!=(e+e−12)(e−e−12)=e4−14e2\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{1}{(2i)!(2j+1)!} = \left(\frac{e+e^{-1}}{2}\right)\left(\frac{e-e^{-1}}{2}\right) = \frac{e^4-1}{4e^2}

P.S. : The answer you have found is absolutely correct.

Therefore, as you see above I haven't used any advanced calculus methods to solve the above sum. Everything lies in just using basic techniques to simplify the sum and then use the results of well-known series.

Thanks,

Kishlaya Jaiswal.

Kishlaya Jaiswal - 6 years, 3 months ago

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Thanks @Kishlaya Jaiswal ! Help is appreciated! :D

Sanchit Ahuja - 6 years, 3 months ago

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It was all mine pleasure in explaining it. Also, I would encourage you to keep posting such interesting notes and feel free to ask for any help. ⌣¨\ddot \smile

Kishlaya Jaiswal - 6 years, 3 months ago

Its correct. You can usually find answers to the Proofathon Contests on the website itself.

Siddhartha Srivastava - 6 years, 3 months ago

May be u should first master the single summation, then go for double summation. I think it needs calculus, and being an 11th class student mr.Ahuja, you are not aware of high level calculus. So try to deviate your focus towards basics first.

ALL THE BEST

rohan bansal - 6 years, 3 months ago

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FYI Mr. Bansal I did single summation and as stated above, it didn't involve any calculus. Thanks Mr. Bansal. For your "Valuable" piece of advice. All the best!

Sanchit Ahuja - 6 years, 3 months ago
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