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____.
Geometric Series Identification
The given series is $1+\frac{1}{3}+\frac{1}{3^2} + \frac{1}{3^3} + ...$. This is an infinite geometric series.
Series Parameters
- First term, $a = 1$.
- Common ratio, $r = \frac{1/3}{1} = \frac{1}{3}$.
Convergence Check
Since the absolute value of the common ratio $|r| = |\frac{1}{3}| < 1$, the series converges to a finite sum.
Sum Calculation
The formula for the sum ($S$) of a converging infinite geometric series is:
$S = \frac{a}{1-r}$
Substitute the values of $a$ and $r$:
$S = \frac{1}{1 - \frac{1}{3}}$
$S = \frac{1}{\frac{3-1}{3}} = \frac{1}{\frac{2}{3}}$
$S = 1 \times \frac{3}{2} = \frac{3}{2}$
$S = 1.5$
Rounding to One Decimal Place
The calculated sum is $1.5$. Rounding this to one decimal place gives $1.5$.
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 __________.
- An infinite series S is given as:
$S = 1 + 2/3 + 3/9 + 4/27 + 5/81 + \dots$ (to infinity)
The value of S is ________________ (round off to 2 decimal places).
