Find convergence of: summation at n=1 going to infinity n^3/5^n
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I assume you know that ∑∞n=0xn=1/(1−x)∑n=0∞xn=1/(1−x)for xx ϵϵ (−1,1)(−1,1).
Then, note that ∑∞n=0xn=1+∑∞n=1xn∑n=0∞xn=1+∑n=1∞xn.
Substituting this in for the left side of the original equation, we can take the derivative of both sides: ddx(1+∑∞n=1xn)=ddx1/(1−x).ddx(1+∑n=1∞xn)=ddx1/(1−x).
Evaluating each derivative gives
0+∑∞n=1nxn−1=1/(1−x)20+∑n=1∞nxn−1=1/(1−x)2
Then we can simplify the inside of the sum, nxn−1nxn−1 into nxn/xnxn/x in order to manipulate the series into 1/x∗∑∞n=1nxn=1/(1−x)21/x∗∑n=1∞nxn=1/(1−x)2. (The 1/x1/x factor is independent of the set of constants nn in the series, so we can factor it out.)
Multiplying through by xx gives ∑∞n=1nxn=x/(1−x)2∑n=1∞nxn=x/(1−x)2.
Finally, we can simply plug in x=1/3x=1/3 to both sides and use the formula on the right side to evaluate the sum!
∑∞n=1n/3n=(1/3)/((1−1/3)2)∑n=1∞n/3n=(1/3)/((1−1/3)2)=(1/3)/(4/9)=3/4(1/3)/(4/9)=3/4.
(This is valid because the original interval of convergence was −1<x<1−1<x<1, so the same is true of the derivative except possibly at the endpoints, though 1/31/3 is within this interval so the endpoints are irrelevant.)
Series are fascinating objects that require lots of creativity to evaluate. Taking derivatives (or antiderivatives) of series often yields interesting results such as that above with relatively few steps. For instance, many series that involve an xnxn and several terms of n(n−1)(n−2)...n(n−1)(n−2)... can be evaluated by taking derivatives of geometric series. Likewise, series with an xnxn and terms of 1/((n+1)(n+2)...)1/((n+1)(n+2)...) can often be evaluated by integrating the geometric series. The geometric series is extremely important in these evaluations, as it is one of few series that can be directly evaluated independently of any prior knowledge of series. Indeed, it even provides the rationale behind the ratio and root tests for convergence, as well as appearing as a component of Taylor series.
Hope that helps!
Then, note that ∑∞n=0xn=1+∑∞n=1xn∑n=0∞xn=1+∑n=1∞xn.
Substituting this in for the left side of the original equation, we can take the derivative of both sides: ddx(1+∑∞n=1xn)=ddx1/(1−x).ddx(1+∑n=1∞xn)=ddx1/(1−x).
Evaluating each derivative gives
0+∑∞n=1nxn−1=1/(1−x)20+∑n=1∞nxn−1=1/(1−x)2
Then we can simplify the inside of the sum, nxn−1nxn−1 into nxn/xnxn/x in order to manipulate the series into 1/x∗∑∞n=1nxn=1/(1−x)21/x∗∑n=1∞nxn=1/(1−x)2. (The 1/x1/x factor is independent of the set of constants nn in the series, so we can factor it out.)
Multiplying through by xx gives ∑∞n=1nxn=x/(1−x)2∑n=1∞nxn=x/(1−x)2.
Finally, we can simply plug in x=1/3x=1/3 to both sides and use the formula on the right side to evaluate the sum!
∑∞n=1n/3n=(1/3)/((1−1/3)2)∑n=1∞n/3n=(1/3)/((1−1/3)2)=(1/3)/(4/9)=3/4(1/3)/(4/9)=3/4.
(This is valid because the original interval of convergence was −1<x<1−1<x<1, so the same is true of the derivative except possibly at the endpoints, though 1/31/3 is within this interval so the endpoints are irrelevant.)
Series are fascinating objects that require lots of creativity to evaluate. Taking derivatives (or antiderivatives) of series often yields interesting results such as that above with relatively few steps. For instance, many series that involve an xnxn and several terms of n(n−1)(n−2)...n(n−1)(n−2)... can be evaluated by taking derivatives of geometric series. Likewise, series with an xnxn and terms of 1/((n+1)(n+2)...)1/((n+1)(n+2)...) can often be evaluated by integrating the geometric series. The geometric series is extremely important in these evaluations, as it is one of few series that can be directly evaluated independently of any prior knowledge of series. Indeed, it even provides the rationale behind the ratio and root tests for convergence, as well as appearing as a component of Taylor series.
Hope that helps!
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limn→∞(54)n1+(34)n=limn →∞(54)n⋅limn→∞11+(34)n=+ ... Then, 5n3n +4n>5n2(4n)
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