The sum of the infinite series

$$
\frac{1}{5} \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \left( \frac{1}{5} \right)^2 - \left( \frac{1}{7} \right)^2 \right) + \left( \frac{1}{5} \right)^3 \left( \left( \frac{1}{5} \right)^3 - \left( \frac{1}{7} \right)^3 \right) + \cdots
$$

is equal to

Question

The sum of the infinite series

$$
\frac{1}{5} \left( \frac{1}{5} - \frac{1}{7} \right) + \left( \frac{1}{5} \right)^2 \left( \left( \frac{1}{5} \right)^2 - \left( \frac{1}{7} \right)^2 \right) + \left( \frac{1}{5} \right)^3 \left( \left( \frac{1}{5} \right)^3 - \left( \frac{1}{7} \right)^3 \right) + \cdots
$$

is equal to

Accepted Answer

the brackets, the series can be arranged as:

$$
\left( \frac{1}{5} ight)^2 - \left( \frac{1}{5} 	imes \frac{1}{7} ight) + \left( \frac{1}{5} ight)^4 - \left( \frac{1}{5} 	imes \frac{1}{7} ight)^2 + \left( \frac{1}{5} ight)^6 - \left( \frac{1}{5} 	imes \frac{1}{7} ight)^3 + ...
$$

$$
= 
\left[ 
\left( 
\frac{1}{5}
ight)^2 +
\left(
\frac{1}{5}
ight)^4 +
\left(
\frac{1}{5}
ight)^6 +
...
ight]
-
\left[
\left(
\frac{1}{5}
	imes
\frac{1}{7}
ight) +
\left(
\frac{1}{5}
	imes
\frac{1}{7}
ight)^2 +
...
ight]
$$

Above is a difference of two geometric series with infinite terms in each.

---

### First Series:

$$
= 
\left(
\frac{1}{5}
ight)^2 +
\left(
\frac{1}{5}
ight)^4 +
\left(
\frac{1}{5}
ight)^6 +
...
$$

Using the formula for the sum of an infinite geometric series:

$$
S = 
a + ar + ar^2 + ...
= 
a / (1 - r)
$$

Here, $$a = \left(\frac{1}{5}ight)^2$$ and $$r = \left(\frac{1}{5}ight)^2$$.

$$
S = 
\frac{\left(\frac{1}{5}ight)^2}
{1 - \left(\frac{1}{5}ight)^2}
= 
\frac{\frac{1}{25}}
{\frac{24}{25}}
= 
\frac{1}{24}.
$$

---

### Second Series:

$$
= 
\left(
\frac{1}{5}
	imes
\frac{1}{7}
ight) +
\left(
\frac{1}{5}
	imes
\frac{1}{7}
ight)^2 +
...
$$

Using the same formula for the sum of an infinite geometric series:

Here, $$a = \left(\frac{1}{5} 	imes \frac{1}{7}ight)$$ and $$r = a$$.

$$
S = 
a / (1 - r)
=
\frac{\left(\frac{1}{35}ight)}
{\left( 1 - \frac{1}{35} ight)}
=
\frac{\frac{1}{35}}
{\frac{34}{35}}
=
\frac{1}{34}.
$$

---

### Final Calculation:

The given series becomes:

$$
S = 
	ext{{First Series}} - 
	ext{{Second Series}}
=
\frac{1}{24} -
\frac{10}{816}.
$$

Simplifying further:

$$
S =
\boxed{\frac{5}{408}}.
$$

Answer

Answer

Answer