Hi,
I am trying to find a closed form for the power series from n=2 to infinity of
x^n/(n-1)n. I'm not really sure how to do this- I found the first and second derivatives which were x^(n-1)/(n-1) and x^n-2. However, this didnt really seem to help much. Any help is appreciated!
To find a closed form expression for the power series, let's break it down step by step.
First, let's rewrite the given series:
S = ∑[from n=2 to ∞] (x^n)/(n-1)n
To simplify the series, we can rewrite the expression x^n as x^(n-1) * x.
S = ∑[from n=2 to ∞] (x^(n-1) * x)/(n-1)n
Now, let's split this into two separate series:
S1 = ∑[from n=2 to ∞] (x^(n-1))/(n-1)
S2 = ∑[from n=2 to ∞] x/(n-1)n
Now we need to find a closed form expression for S1 and S2 separately.
For S1, let's notice that it resembles the derivative of a well-known function. Taking the derivative of x^n with respect to x gives us nx^(n-1). So, we can rewrite S1 as:
S1 = ∑[from n=2 to ∞](d/dx(x^n))/(n-1)
Now, take the derivative of both sides of S1 with respect to x:
dS1/dx = ∑[from n=2 to ∞] (d/dx(d/dx(x^n)))/(n-1) = ∑[from n=2 to ∞] (d^2/dx^2)(x^n)/(n-1)
To find the closed form expression for dS1/dx, let's express it in terms of another familiar power series. Let's define a new series T as:
T = ∑[from n=1 to ∞] (x^n)/(n)
Now, let's find the second derivative of T with respect to x:
(d^2/dx^2)T = (d^2/dx^2)∑[from n=1 to ∞] (x^n)/(n) = ∑[from n=1 to ∞] (d^2/dx^2)(x^n)/(n)
Comparing this with our expression for dS1/dx, we can conclude that:
(d^2/dx^2)T = dS1/dx
Integrating (d^2/dx^2)T with respect to x twice will give us T, which is a well-known power series for the natural logarithm of (1 + x):
T = ln(1 + x)
Therefore, the closed form expression for S1 is given by:
S1 = ∫∫(d^2/dx^2)T dx = ∫∫(d/dx)dS1/dx dx = ∫∫dS1 dx = ∫S1 dx = ∫ln(1 + x) dx
Now, let's find the closed form expression for S2.
For S2, we can rewrite it as:
S2 = x * ∑[from n=2 to ∞](1/(n-1)n)
This series can be simplified using partial fractions. Let's rewrite the term 1/(n-1)n as A/(n-1) + B/n, where A and B are constants to be determined.
Combining the fractions, we get:
1/(n-1)n = (A(n) + B(n-1))/((n-1)n)
Now, let's find the values of A and B. Multiply both sides by (n-1)n to eliminate the denominator:
1 = (A(n) + B(n-1))
For n = 2, we have:
1 = (2A + B)
For n = 3, we have:
1 = (3A + 2B)
Solving these two equations simultaneously will give us the values of A and B.
Once we have A and B, we can rewrite S2 as:
S2 = x * ∑[from n=2 to ∞](A/(n-1) + B/n)
Now, let's find the closed form expression for S2 by evaluating the series:
S2 = x * ∑[from n=2 to ∞](A/(n-1) + B/n) = x * (A * (∑[from n=2 to ∞]1/(n-1)) + B * (∑[from n=2 to ∞]1/n))
The series ∑[from n=2 to ∞]1/(n-1) is a telescoping series that simplifies to ∑[from n=2 to ∞]1/n.
Therefore, the closed form expression for S2 is given by:
S2 = x * (A * ln(1) + B * (∑[from n=2 to ∞]1/n)) = x * (B * (∑[from n=2 to ∞]1/n)) = x * (B * [ln(n)]|[from n=2 to ∞])
Now, we have the closed form expressions for both S1 and S2. By adding these two series together, we can find the closed form expression for the original series S.
I hope this explanation helps! If you have any further questions, please let me know.