Advanced Series

Here is a fun infinite series problem I made. I originally posted this problem on Brilliant.org.

## **Problem**

Consider the sum and its incomplete solution:

∑

# n
2

∞

e

−
2
n

n
2

−
1

=

1
A

(

1

e

−
B

C

)

sinh
⁡
(
D
)
ln
⁡

(

1
−

e

−
E

)

.

{\displaystyle \sum\_{n=2}^{\infty} \frac{e^{-2n}}{n^2 - 1} = \frac{1}{A}\left(1 + \frac{e^{-B}}{C}\right) + \sinh(D) \ln \left(1-e^{-E}\right). }

Determine the product

A
B
C
D
E

{\displaystyle ABCDE}

.

## **Hint**

(1) Use the infinite series

−
ln
⁡
(
1
−
x

# n
1

∞

x
n

n

.

{\displaystyle -\ln(1-x) = \sum\_{n=1}^{\infty} \frac{x^n}{n}.}

## **Solution**

Factorize the denominator of the series and decompose this series into partial fractions.

∑

# n
2

∞

e

−
2
n

n
2

−
1

=

∑

# n
2

∞

e

−
2
n

(
n
−
1
)
(
n
1
)

{\displaystyle \sum\_{n=2}^{\infty} \frac{e^{-2n}}{n^2 - 1} = \sum\_{n=2}^{\infty} \frac{e^{-2n}}{(n-1)(n+1)} }

∑

# n
2

∞

e

−
2
n

n
2

−
1

=

1
2

(

∑

# n
2

∞

e

−
2
n

n
−
1

−

∑

# n
2

∞

e

−
2
n

n
1

)

.

{\displaystyle \sum\_{n=2}^{\infty} \frac{e^{-2n}}{n^2 - 1} = \frac{1}{2}\left(\sum\_{n=2}^{\infty} \frac{e^{-2n}}{n-1} - \sum\_{n=2}^{\infty} \frac{e^{-2n}}{n+1} \right). }

Write out the first three terms of each decomposed series

1
2

[

(

e

−
4

e

−
6

2

e

−
8

3

.
.
.

)

−

(

e

−
4

3

e

−
6

4

e

−
8

5

.
.
.

)

]

{\displaystyle \frac{1}{2}\left[\left(e^{-4} + \frac{e^{-6} }{2} + \frac{e^{-8} }{3}+...\right) - \left(\frac{e^{-4}}{3}+ \frac{e^{-6}}{4}+ \frac{e^{-8}}{5}+...\right) \right]}

and factor out

e

−
2

{\displaystyle e^{-2} }

e

−
2

2

[

(

e

−
2

e

−
4

2

e

−
6

3

.
.
.

)

−

(

e

−
2

3

e

−
4

4

e

−
6

5

.
.
.

)

]

.

{\displaystyle \frac{e^{-2}}{2}\left[\left(e^{-2} + \frac{e^{-4} }{2} + \frac{e^{-6} }{3}+...\right) - \left(\frac{e^{-2}}{3}+ \frac{e^{-4}}{4}+ \frac{e^{-6}}{5}+...\right) \right].}

Use the Taylor series

−
ln
⁡
(
1
−
x

# n
1

∞

x
n

n

{\displaystyle -\ln(1-x) = \sum\_{n=1}^{\infty} \frac{x^n}{n}}

−
ln
⁡
(
1
−
x
# )
x

x
2

2

x
3

3

.
.
.

{\displaystyle -\ln(1-x) = x+\frac{x^2}{2}+\frac{x^3}{3}+... }

.
to rewrite the series as

e

−
2

e

−
4

2

e

−
6

3

.
.
# .
−
ln
⁡
(
1
−

e

−
2

)
.

{\displaystyle e^{-2} + \frac{e^{-4} }{2} + \frac{e^{-6} }{3}+... = -\ln(1-e^{-2}).}

Observe that

e

−
2

3

e

−
4

4

e

−
6

5

.
.
# .

(

e

−
2

e

−
4

2

e

−
6

3

.
.
.
−

e

−
2

−

e

−
4

2

)

e
4

.

{\displaystyle \frac{e^{-2}}{3}+ \frac{e^{-4}}{4}+ \frac{e^{-6}}{5}+... = \left(e^{-2} + \frac{e^{-4}}{2} + \frac{e^{-6} }{3}+...-e^{-2} - \frac{e^{-4}}{2} \right)e^4. }

Therefore,

e

−
2

3

e

−
4

4

e

−
6

5

.
.
# .
−

e
4

ln
⁡
(
1
−

e

−
2

)
−

e
2

−

1
2

.

{\displaystyle \frac{e^{-2}}{3}+ \frac{e^{-4}}{4}+ \frac{e^{-6}}{5}+... = -e^4\ln(1-e^{-2}) - e^2 - \frac{1}{2}. }

Substitute the above compact expressions of the infinite series.

∑

# n
2

∞

e

−
2
n

n
2

−
1

=

e

−
2

2

[

−
ln
⁡
(
1
−

e

−
2

)
−

(

−

e
4

ln
⁡
(
1
−

e

−
2

)
−

e
2

−

1
2

)

]

=

e
2

ln
⁡
(
1
−

e

−
2

)

2

−

e

−
2

ln
⁡
(
1
−

e

−
2

)

2

1
2

e

−
2

4

=

1
2

(

1

e

−
2

2

)

(

e

2

−

e

−
2

2

)

ln
⁡
(
1
−

e

−
2

)

=

1
2

(

1

e

−
2

2

)

sinh
⁡
(
2
)
ln
⁡
(
1
−

e

−
2

)
.

{\displaystyle \begin{align}
\sum\_{n=2}^{\infty} \frac{e^{-2n}}{n^2 - 1} &= \frac{e^{-2}}{2}\left[-\ln(1-e^{-2})-\left(-e^4\ln(1-e^{-2}) - e^2 - \frac{1}{2}\right) \right] \\[5 pt]
&= \frac{e^2 \ln(1-e^{-2})}{2}- \frac{e^{-2} \ln(1-e^{-2})}{2} + \frac{1}{2} + \frac{e^{-2}}{4} \\[5 pt]
&= \frac{1}{2}\left(1+\frac{e^{-2}}{2} \right)+\left(\frac{e^{2} - e^{-2}}{2}\right) \ln(1-e^{-2}) \\[5 pt]
&= \frac{1}{2}\left(1+\frac{e^{-2}}{2} \right)+\sinh(2) \ln(1-e^{-2}).
\end{align} }

This means

# E
2

{\displaystyle A = B = C = D = E = 2 }

 and the product

A
B
C
D
# E
32

{\displaystyle ABCDE = 32 }

.