To find the value of the double series $\sum_{i=0}^{\infty} \sum_{j=1}^{\infty} 2^{-i} 3^{-j}$, we start by calculating the two inner and outer sums separately. The double series can be rewritten as: \[ \sum_{i=0}^{\infty} \left( \sum_{j=1}^{\infty} 2^{-i} 3^{-j} \right) \]
First, consider the inner sum: \[ \sum_{j=1}^{\infty} 3^{-j} \]
This is a geometric series where the first term \(a=3^{-1}=\frac{1}{3}\) and the common ratio \(r=3^{-1}=\frac{1}{3}\).
The sum of an infinite geometric series is given by \( \frac{a}{1-r} \). Plugging in the values, we have: \[ \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]
Next, substitute this result back into the outer sum: \[ \sum_{i=0}^{\infty} 2^{-i} \cdot \frac{1}{2} \]
This can be expressed as: \[ \frac{1}{2} \sum_{i=0}^{\infty} 2^{-i} \]
Now, consider the series for \( \sum_{i=0}^{\infty} 2^{-i} \), another geometric series with \( a=1 \) and \( r=2^{-1}=\frac{1}{2} \).
The sum is: \[ \frac{1}{1-\frac{1}{2}}=2 \]
Thus, substituting back: \[ \frac{1}{2} \cdot 2 = 1 \]
We find that the value of the double series is \( 1 \), which lies within the given range of 1 to 1.
Important Questions from Infinite Series
- Consider the following series:
(i) $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
(ii) $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$
(iii) $\sum_{n=1}^{\infty} \frac{1}{n!}$ - The sum of the following infinite series is:
$ \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + ... $ - The series
$\sum_{n=0}^{r} q^n = 1 + q + q^2 + \dots$ has the sum: The value of the series $1+ \sin x + \cos^2 x + \sin^3 x + \dots$ at $x = \frac{ \pi}{4}$ is __________.
The sum of the infinite geometric series $1+\frac{1}{3}+\frac{1}{3^2} + \frac{1}{3^3} + ...$ (rounded off to one decimal place) is____.
