Mister Exam
Sum of series 1/((1+r)^n)
The solution
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oo ____ \ ` \ 1 \ -------- / n / (1 + r) /___, n = 1
$$\sum_{n=1}^{\infty} \frac{1}{\left(r + 1\right)^{n}}$$
Sum(1/((1 + r)^n), (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{1}{\left(r + 1\right)^{n}}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = 1$$
and
$$x_{0} = - r - 1$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(- r - 1 + \lim_{n \to \infty} 1\right)$$
Let's take the limit
we find
$$\frac{1}{R} = \tilde{\infty} r$$
$$R = 0$$
$$\frac{1}{\left(r + 1\right)^{n}}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = 1$$
and
$$x_{0} = - r - 1$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(- r - 1 + \lim_{n \to \infty} 1\right)$$
Let's take the limit
we find
$$\frac{1}{R} = \tilde{\infty} r$$
$$R = 0$$
The answer
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/ 1 1 |------------------- for ------- < 1 | / 1 \ |1 + r| |(1 + r)*|1 - -----| | \ 1 + r/ | < oo | ___ | \ ` | \ -n | / (1 + r) otherwise | /__, \ n = 1
$$\begin{cases} \frac{1}{\left(1 - \frac{1}{r + 1}\right) \left(r + 1\right)} & \text{for}\: \frac{1}{\left|{r + 1}\right|} < 1 \\\sum_{n=1}^{\infty} \left(r + 1\right)^{- n} & \text{otherwise} \end{cases}$$
Piecewise((1/((1 + r)*(1 - 1/(1 + r))), 1/|1 + r| < 1), (Sum((1 + r)^(-n), (n, 1, oo)), True))
Examples of finding the sum of a series
- The Sum of the Power Series
x^n/n
(x-1)^n
- Factorial
1/2^(n!)
n^2/n!
x^n/n!
k!/(n!*(n+k)!)
- Flint Hills Series
csc(n)^2/n^3
- Basel problem
1/n^2
1/n^4
1/n^6
- Harmonic series
1/n
- Grandi's series
(-1)^n
- Alternating series
(-1)^(n + 1)/n
(n + 2)*(-1)^(n - 1)
(3*n - 1)/(-5)^n
(-1)^(n - 1)*n/(6*n - 5)
- Newton–Mercator series
(-1)^(n + 1)/n*x^n
- Exam the series for convergence
(3*n - 1)/(-5)^n