[sp]Differentiate the geometric series $1 + k + k^2 + \ldots + k^{n+1} = \dfrac{1-k^{n+2}}{1-k}$ with respect to $k$, to get $$1 + 2k + 3k^2 + \ldots + (n+1)k^n = \frac{-(n+2)k^{n+1}(1-k) + 1 - k^{n+2}}{(1-k)^2} = \frac{(n+1)k^{n+2} - (n+2)k^{n+1} + 1}{(1-k)^2} \to \frac1{(1-k)^2}$$ as $n \to \infty$.lfdahl said:Let $S_n(k)$ be defined by:
$S_n(k) = 1 + 2k+3k^2+...+(n+1)k^n$, where $|k| < 1$ and $n \in \Bbb{N}$.
Derive a closed form for $S_n(k)$ and find the limit: $$\lim_{{n}\to{\infty}}S_n(k).$$
Opalg said:[sp]Differentiate the geometric series $1 + k + k^2 + \ldots + k^{n+1} = \dfrac{1-k^{n+2}}{1-k}$ with respect to $k$, to get $$1 + 2k + 3k^2 + \ldots + (n+1)k^n = \frac{-(n+2)k^{n+1}(1-k) + 1 - k^{n+2}}{(1-k)^2} = \frac{(n+1)k^{n+2} - (n+2)k^{n+1} + 1}{(1-k)^2} \to \frac1{(1-k)^2}$$ as $n \to \infty$.
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MarkFL said:My solution:
We may express $S_n$ in the following difference equation:
\(\displaystyle S_{n}-S_{n-1}=(n+1)k^n\)
The homogeneous solution is:
\(\displaystyle h_n=c_1\)
And the particular solution will take the form:
\(\displaystyle p_n=\left(c_2n+c_3\right)k^n\)
Substitute into our difference equation:
\(\displaystyle \left(c_2n+c_3\right)k^n-\left(c_2(n-1)+c_3\right)k^{n-1}=(n+1)k^n\)
Divide through by $k^{n-1}$:
\(\displaystyle \left(c_2n+c_3\right)k-\left(c_2(n-1)+c_3\right)=(n+1)k\)
Arrange as:
\(\displaystyle c_2(k-1)n+\left(c_3(k-1)+c_2\right)=kn+k\)
Equating like coefficients yields the system:
\(\displaystyle c_2(k-1)=k\implies c_2=-\frac{k}{1-k}\)
\(\displaystyle c_3(k-1)+c_2=c_3(k-1)-\frac{k}{1-k}=k\implies c_3=-\frac{k(2-k)}{(1-k)^2}\)
And thus our particular solution is:
\(\displaystyle p_n=-\left(\frac{k}{1-k}n+\frac{k(2-k)}{(1-k)^2}\right)k^n=-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}\)
And so by the principle of superposition, we have:
\(\displaystyle S_n=p_n+h_n=-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}+c_1\)
Using the initial value (extending $n$ to $\mathbb{N_0}$ for simplicity), we find:
\(\displaystyle S_0=-\left(\frac{2-k}{(1-k)^2}\right)k+c_1=1\implies c_1=\frac{1}{(1-k)^2}\)
And so the solution satisfying the given conditions is:
\(\displaystyle S_n=\frac{1}{(1-k)^2}-\left(\frac{(1-k)n+2-k}{(1-k)^2}\right)k^{n+1}=\frac{1-\left((1-k)n+2-k\right)k^{n+1}}{(1-k)^2}\)
As \(\displaystyle \lim_{n\to\infty}\left(\frac{an+b}{c^n}\right)=0\) for $1<c$, we see that:
\(\displaystyle S_{\infty}=\lim_{n\to\infty}S_n=\frac{1}{(1-k)^2}\)
Euge said:Here is my solution.
Note \begin{align}S_n(k) - kS_n(k) &= 1 + (2k - k) + (3k^2 - 2k^2) + \cdots + [(n+1)k^n - nk^n] - (n+1)k^{n+1}\\
(1-k)S_n(k) &= 1 + k + k^2 + \cdots + k^n - (n+1)k^{n+1}\\
(1-k)S_n(k) &= \frac{1-k^{n+1}}{1-k} - (n+1)k^{n+1}\end{align}
Since $k^{n+1}$ and $(n+1)k^{n+1}$ tend to $0$ as $n\to \infty$ (since $\lvert k\rvert < 1$), then setting $L = \lim\limits_{n\to \infty} S_n(k)$, one obtains $(1-k)L = \dfrac{1}{1-k}$, or $L = \dfrac{1}{(1-k)^2}$.
Deriving a closed form refers to finding a mathematical expression that represents the solution to a problem in terms of a finite number of basic arithmetic operations. This allows for a more concise and general solution compared to using numerical methods.
A closed form can be derived by using techniques such as algebraic manipulation, differentiation, and integration to transform a given problem into a simpler, more compact form. This is often done in mathematics and physics to find exact solutions to problems.
The limit of a closed form is the value that the expression approaches as the independent variable tends towards a particular value or infinity. It is often denoted by the symbol "lim" and is used to describe the behavior of the expression as a whole.
Finding the limit of a closed form is important because it allows us to understand the behavior of a mathematical expression as the input values change. This can help us make predictions and draw conclusions about the relationship between variables in a problem.
No, not all problems can be solved using a closed form and limit approach. Some problems may require numerical methods or other techniques to find a solution, while others may not have a closed form solution at all. It ultimately depends on the complexity and nature of the problem.