Telescoping Series 1

## **Problem**

Calculate the infinite series

3

1
×
2
×
3

5

2
×
3
×
4

7

3
×
4
×
5

9

4
×
5
×
6

⋯
.

{\displaystyle \frac{3}{1\times2\times3} + \frac{5}{2\times3\times4} + \frac{7}{3\times4\times5} + \frac{9}{4\times5\times6} + \cdots . }

## **Hints**

(1) Using summation notation, the infinite series is

∑

# n
1

∞

2
n
1

n
(
n
1
)
(
n
2
)

.

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

(2) Use partial fractions; this is an essential technique for evaluating telescoping series.

## **Solution**

Write the infinite series using summation notation:

∑

# n
1

∞

2
n
1

n
(
n
1
)
(
n
2
)

.

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

Decompose the pattern by partial fractions

2
n
1

n
(
n
1
)
(
n
2
)

=

A
n

B

n
1

C

n
2

{\displaystyle \frac{2n+1}{n(n+1)(n+2)} = \frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} }

where the coefficients

A
,
B
,
C

{\displaystyle A, B, C}

are real numbers.

Thus,

2
n
1

n
(
n
1
)
(
n
2
)

=

A
(
n
1
)
(
n
2
)
B
(
n
)
(
n
2
)
C
(
n
)
(
n
1
)

n
(
n
1
)
(
n
2
)

{\displaystyle \frac{2n+1}{n(n+1)(n+2)} = \frac{A(n+1)(n+2) + B(n)(n+2)+C(n)(n+1)}{n(n+1)(n+2)} }

By inspecting the numerator, we get

2
n
# 1
A
(
n
1
)
(
n
2
)
B
(
n
)
(
n
2
)
C
(
n
)
(
n
1
)
.

{\displaystyle 2n+1 = A(n+1)(n+2) + B(n)(n+2)+C(n)(n+1). }

Setting

# n
0

{\displaystyle n=0}

, we get

# 1
2
A

{\displaystyle 1 = 2A }

.

Setting

# n
−
1

{\displaystyle n = -1}

, we get

−
# 1
−
B

{\displaystyle -1 = -B }

.

Setting

# n
−
2

{\displaystyle n=-2 }

, we get

−
# 3
2
A

{\displaystyle -3 = 2A }

.

Solving for each coefficient yields:

{

A

=

1
2

B

=
1

C

=

−
3

2

.

{\displaystyle \begin{cases} A &= \frac{1}{2} \\ B &= 1 \\ C &= \frac{-3}{2}. \end{cases} }

Substituting the solved coefficients into the *n*th partial sum of the infinite series, we get

S
n

=

(

1

/

2

1

1
2

−

3

/

2

3

)

(

1

/

2

2

1
3

−

3

/

2

4

)

(

1

/

2

3

1
4

−

3

/

2

5

)

⋯

(

1

/

2

n

1

n
1

−

3

/

2

n
2

)

.

{\displaystyle S\_n = \left(\frac{1/2}{1} + \frac{1}{2} -\frac{3/2}{3}\right) + \left(\frac{1/2}{2} + \frac{1}{3} -\frac{3/2}{4}\right) + \left(\frac{1/2}{3} + \frac{1}{4} -\frac{3/2}{5}\right) + \cdots + \left(\frac{1/2}{n} + \frac{1}{n+1} -\frac{3/2}{n +2}\right). }

For this problem, the telescoping takes place across groupings. The last term of one triple cancels with the sum of the middle term of the next grouping and the first term of the third grouping, which is equivalent to the equation:

−

3

/

2

n

1
n

1

/

2

n

=

{\displaystyle -\frac{3/2}{n} + \frac{1}{n} + \frac{1/2}{n} = 0. }

Thus, the *n*th partial sum can be rewritten as

S
n

=

(

1

/

2

1

1
2

)

1

/

2

2

−

3

/

2

n
1

(

1

n
1

−

3

/

2

n
2

)

=

5
4

−

1

/

2

n
1

−

3

/

2

n
2

.

{\displaystyle \begin{align}
S\_n &= \left(\frac{1/2}{1} + \frac{1}{2} \right) + \frac{1/2}{2} - \frac{3/2}{n+1} + \left(\frac{1}{n+1} -\frac{3/2}{n+2}\right) \\
&= \frac{5}{4} - \frac{1/2}{n+1} - \frac{3/2}{n+2}.
\end{align} }

To evaluate the infinite series, take the limit to infinity:

∑

# n
1

∞

2
n
1

n
(
n
1
)
(
n
2
)

=

lim

n
→
∞

5
4

−

1

/

2

n
1

−

3

/

2

n
2

=

5
4

.

{\displaystyle \begin{align}
\sum\_{n=1}^\infty \frac{2n+1}{n(n+1)(n+2)} &= \lim\_{n \rightarrow \infty} \frac{5}{4} - \frac{1/2}{n+1} - \frac{3/2}{n+2} \\
&= \frac{5}{4}.
\end{align} }

Therefore,

3

1
×
2
×
3

5

2
×
3
×
4

7

3
×
4
×
5

9

4
×
5
×
6

# ⋯

5
4

.

{\displaystyle \frac{3}{1\times2\times3} + \frac{5}{2\times3\times4} + \frac{7}{3\times4\times5} + \frac{9}{4\times5\times6} + \cdots = \frac{5}{4}. }