Sum of infinite series $\frac{-5}{4},\frac{5}{16},\frac{-5}{64},......$ is

Question

Accepted Answer

The given series is an infinite geometric series with the first term $ a = \frac{-5}{4} $ and the common ratio $ r $. To find the common ratio, we divide the second term by the first term:

$$
r = \frac{\frac{5}{16}}{\frac{-5}{4}} = \frac{5}{16} \times \frac{-4}{5} = -\frac{1}{4}
$$

The formula for the sum of an infinite geometric series is given by:

$$
S = \frac{a}{1 - r}
$$

This formula is applicable only if the absolute value of the common ratio $ |r| < 1 $. In this case, $ |r| = \frac{1}{4} $, which is less than 1, so the formula can be used.

Substituting the values of $ a $ and $ r $ into the formula:

$$
S = \frac{\frac{-5}{4}}{1 - \left(-\frac{1}{4}\right)} = \frac{\frac{-5}{4}}{1 + \frac{1}{4}} = \frac{\frac{-5}{4}}{\frac{5}{4}}
$$

Simplifying the expression:

$$
S = \frac{-5}{4} \times \frac{4}{5} = -1
$$

Thus, the sum of the infinite series is $-1$, which corresponds to Option 2.

**Explanation for Incorrect Options:**

1. **Option 1 (1):** This option suggests that the sum of the series is 1. However, as calculated, the sum is $-1$, not 1. The calculation clearly shows that the series converges to $-1$.

2. **Option 3 ($\infty$):** This option implies that the series diverges to positive infinity. For a geometric series to diverge, the absolute value of the common ratio must be greater than or equal to 1. Since $|r| = \frac{1}{4}$, which is less than 1, the series converges rather than diverges.

3. **Option 4 ($-\infty$):** This option implies that the series diverges to negative infinity. Similar to the reasoning for Option 3, the series converges because $|r| = \frac{1}{4}$ is less than 1. Therefore, it cannot diverge to negative infinity.

In conclusion, the correct answer is Option 2, as the sum of the infinite series is indeed $-1$.