$S = 1 + 2/3 + 3/9 + 4/27 + 5/81 + \dots$ (to infinity)
The value of S is ________________ (round off to 2 decimal places).
Series Sum Calculation
The given infinite series is:
$S = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27} + \frac{5}{81} + \dots$
This is an arithmetic-geometric series. We can rewrite it using a common ratio $x = 1/3$:
$S = 1 \cdot x^0 + 2 \cdot x^1 + 3 \cdot x^2 + 4 \cdot x^3 + \dots$
Let's find the sum using the following method:
- Represent the series as $S = \sum_{n=1}^{\infty} n x^{n-1}$.
- Multiply the series by $x$: $xS = \sum_{n=1}^{\infty} n x^{n}$.
- Subtract $xS$ from $S$:
$S - xS = (1 + 2x + 3x^2 + 4x^3 + \dots) - (x + 2x^2 + 3x^3 + \dots)$
$S(1-x) = 1 + x + x^2 + x^3 + \dots$
- The right side is an infinite geometric series $\sum_{n=0}^{\infty} x^n$ with first term $a=1$ and common ratio $x$. Since $|x| = |1/3| < 1$, the sum converges to $\frac{1}{1-x}$.
- Therefore, $S(1-x) = \frac{1}{1-x}$.
- Solve for $S$: $S = \frac{1}{(1-x)^2}$.
Substituting the Value of x
Substitute $x = 1/3$ into the formula for $S$:
$S = \frac{1}{(1 - 1/3)^2}$
$S = \frac{1}{(2/3)^2}$
$S = \frac{1}{4/9}$
$S = \frac{9}{4}$
Final Result
Convert the fraction to a decimal:
$S = 2.25$
The calculated value $S = 2.25$ lies between 2.2 and 2.3, consistent with the problem statement.
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____.
