Mister Exam
Other calculators
- Taylor Series Step by Step
- Fourier series step by step
- Differential equations Step by Step
- Product of series
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How to use it?
- Sum of series:
-
(-1)^n
-
1/(n(n+3))
-
(3^n-4^n)/12^n
-
1/((n+14)(n+15))
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Identical expressions
- (x+ three)^n/ two ^nsqrt(n^ four + four) one /n^ two
- (x plus 3) to the power of n divide by 2 to the power of n square root of (n to the power of 4 plus 4)1 divide by n squared
- (x plus three) to the power of n divide by two to the power of n square root of (n to the power of four plus four) one divide by n to the power of two
- (x+3)^n/2^n√(n^4+4)1/n^2
- (x+3)n/2nsqrt(n4+4)1/n2
- x+3n/2nsqrtn4+41/n2
- (x+3)^n/2^nsqrt(n⁴+4)1/n²
- (x+3) to the power of n/2 to the power of nsqrt(n to the power of 4+4)1/n to the power of 2
- x+3^n/2^nsqrtn^4+41/n^2
- (x+3)^n divide by 2^nsqrt(n^4+4)1 divide by n^2
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Similar expressions
- (x-3)^n/2^nsqrt(n^4+4)1/n^2
- (x+3)^n/2^nsqrt(n^4-4)1/n^2
Sum of series (x+3)^n/2^nsqrt(n^4+4)1/n^2
The solution
You have entered
[src]
oo
______
\ `
\ n ________
\ (x + 3) / 4
\ --------*\/ n + 4
\ n
/ 2
/ --------------------
/ 2
/ n
/_____,
n = 1
$$\sum_{n=1}^{\infty} \frac{\frac{\left(x + 3\right)^{n}}{2^{n}} \sqrt{n^{4} + 4}}{n^{2}}$$
Sum((((x + 3)^n/2^n)*sqrt(n^4 + 4))/n^2, (n, 1, oo))
The radius of convergence of the power series
Given number:
$$\frac{\frac{\left(x + 3\right)^{n}}{2^{n}} \sqrt{n^{4} + 4}}{n^{2}}$$
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
$$a_{n} = \frac{2^{- n} \sqrt{n^{4} + 4}}{n^{2}}$$
and
$$x_{0} = -3$$
,
$$d = 1$$
,
$$c = 1$$
then
$$R = -3 + \lim_{n \to \infty}\left(\frac{2^{- n} 2^{n + 1} \left(n + 1\right)^{2} \sqrt{n^{4} + 4}}{n^{2} \sqrt{\left(n + 1\right)^{4} + 4}}\right)$$
Let's take the limit
we find
$$R = -1$$
$$\frac{\frac{\left(x + 3\right)^{n}}{2^{n}} \sqrt{n^{4} + 4}}{n^{2}}$$
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
$$a_{n} = \frac{2^{- n} \sqrt{n^{4} + 4}}{n^{2}}$$
and
$$x_{0} = -3$$
,
$$d = 1$$
,
$$c = 1$$
then
$$R = -3 + \lim_{n \to \infty}\left(\frac{2^{- n} 2^{n + 1} \left(n + 1\right)^{2} \sqrt{n^{4} + 4}}{n^{2} \sqrt{\left(n + 1\right)^{4} + 4}}\right)$$
Let's take the limit
we find
$$R = -1$$
The answer
[src]
oo _____ \ ` \ ________ \ -n n / 4 \ 2 *(3 + x) *\/ 4 + n / ------------------------ / 2 / n /____, n = 1
$$\sum_{n=1}^{\infty} \frac{2^{- n} \sqrt{n^{4} + 4} \left(x + 3\right)^{n}}{n^{2}}$$
Sum(2^(-n)*(3 + x)^n*sqrt(4 + n^4)/n^2, (n, 1, oo))
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