Somersaults are more fun than this sum assault

Calculus Level 2

∑ n = 0 ∞ 1 ( 4 n ) ! \sum_{n=0}^{\infty} \frac{1}{(4n)!}

The above sum can be expressed in the form

1 n ( ∑ k = 1 m n e i k ) \frac{1}{n} \left( \sum_{k=1}^{mn} e^{i^{k}} \right)

where i i is the imaginary unit and n n is some positive multiple of 4.

Find m m .


The answer is 1.

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Jake Lai by Jake Lai , 21, Singapore

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5 solutions

Recall from Taylor expansion that: c o s x = ∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! x 2 n = 1 − x 2 2 ! + x 4 4 ! − ⋯ cosx = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+ \frac{x^4}{4!}-\cdots c o s h x = ∑ n = 0 ∞ 1 ( 2 n ) ! x 2 n = 1 + x 2 2 ! + x 4 4 ! + ⋯ . coshx = \sum_{n=0}^\infty \frac{1}{(2n)!}x^{2n}=1+\frac{x^2}{2!}+ \frac{x^4}{4!}+\cdots. Now, from Euler's identity c o s x = e i x + e − i x 2 cosx = \frac{e^{ix}+e^{-ix}}{2} and the definition of c o s h = e x + e − x 2 cosh=\frac{e^x+e^{-x}}{2} we can see the problem as c o s ( 1 ) + c o s h ( 1 ) 2 = e i + e − i + e 1 + e − 1 4 = ∑ n = 0 ∞ 1 ( 4 n ) ! . \begin{aligned} \frac{cos(1) +cosh(1)}{2} &= \frac{e^{i}+e^{-i}+e^{1}+e^{-1}}{4} \\ &=\sum_{n=0}^\infty \frac{1}{(4n)!} . \end{aligned}

Finally, recall that i i has period the result follows.

9 Helpful 5 Interesting 3 Brilliant 0 Confused
Otto Bretscher
May 5, 2015

I think the answer is actually m = 1 m=1 (that was my first try!), unless I'm reading the problem wrong.

Using basic Taylor series, we have ∑ k = 0 ∞ 1 ( 4 n ) ! = 1 4 ( 2 cos ⁡ 1 + e + e − 1 ) = 1 4 ( e i + e − i + e + e − 1 ) = 1 4 ∑ k = 1 4 e i k \sum_{k=0}^{\infty}\frac{1}{(4n)!}=\frac{1}{4}(2\cos{1}+e+e^{-1})=\frac{1}{4}(e^i+e^{-i}+e+e^{-1})=\frac{1}{4}\sum_{k=1}^{4}e^{i^k}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Because the periodicity of i n i^n is 4. Your last summation can further be generalize to 1 4 L ∑ k = 1 4 L e i k \displaystyle \frac {1}{4L} \sum_{k=1}^{4L} e^{i^k} .

Pi Han Goh - 6 years, 1 month ago

Yes Answer should be 1 .

by using expansion of e x e^x and replacing

x = > i x x=>ix

x = > − i x x=>-ix

x = > − x x=>-x

x = > + x x=>+x

and then add and devide by 4

I lost my three chances . Since I answer as

1 then 4 then 16 , but never thought to answer 2 .. :P !

But answer should be 1 !

Nishu sharma - 6 years, 1 month ago

Wow!A wonderful problem was this. As the numerator 1 is i to the any 4n I just wrote down the serieses of cosh i and cos i and those remaining even powered terms just cancelled out immediately. A little confusion was due to the dummy n of the first summation was redefined again in the second summation and instead of different mn and n only a single variable can be used as m determines how many coshi+cosi are there so that the form of the second summation will remain working for any m>1.

MAMUNUR RASHID - 6 years, 1 month ago
Abu Zubair
May 8, 2015

amazing question!!! method for beginners :p

3 Helpful 0 Interesting 1 Brilliant 0 Confused
Saúl Huerta
Aug 8, 2020

We observe that:

e i = ∑ k = 0 ∞ i k k ! = 1 + i − 1 2 − i 6 + 1 24 + . . . e^i=\sum_{k=0}^{\infty}\dfrac{i^k}{k!}=1+i-\frac{1}{2}-\frac{i}{6}+\frac{1}{24}+... e − 1 = ∑ k = 0 ∞ ( − 1 ) k k ! = 1 − 1 + 1 2 − 1 6 + 1 24 − . . . e^{-1}=\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k!}=1-1+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-... e − i = ∑ k = 0 ∞ ( − i ) k k ! = 1 − i − 1 2 + i 6 + 1 24 − . . . e^{-i}=\sum_{k=0}^{\infty}\dfrac{(-i)^k}{k!}=1-i-\frac{1}{2}+\frac{i}{6}+\frac{1}{24}-... e = ∑ k = 0 ∞ 1 k ! = 1 + 1 + 1 2 + 1 6 + 1 24 + . . . e=\sum_{k=0}^{\infty}\dfrac{1}{k!}=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+...

Because the periodicity of the powers of i is of 4 and the periodicity of the powers of -1 is of 2, we will get the factorials of the multiples of four with the same positive sign in all the taylor series above. We can also observe that the other terms cancel perfectly. So adding them will yield the desired sum repeated for times so:

∑ n = 0 ∞ 1 ( 4 n ) ! = 1 4 ∑ j = 1 4 e i k = e 2 + 2 e cos ⁡ ( 1 ) + 1 4 e \sum_{n=0}^{\infty}\frac{1}{(4n)!}=\frac{1}{4}\sum_{j=1}^{4}e^{i^k}=\frac{e^{2}+2e\cos(1)+1}{4e} ⟹ m = 1 \implies\boxed{m=1}

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Bogdan Simeonov
May 7, 2015

In fact, this can be generalized for every number using the roots of unity.The sum of the reciprocals of the factorials of the type (nm)!, where n is fixed and m ranges from 1 to infinity, is actually the sum of e to the power if every nth root of unity, divided by n (or just to the power of the powers of a primitive root).In this case, a primitive fourth root of unity is i.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Yup, I noticed that. I was going to post a note on it but I'm hella lazy so.

Jake Lai - 6 years ago

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