Use Theorem 9.4.3 to find the sum of the series.
∞
Σ ( (1/ 5^k) - (1/ (k (k+1) ) )
k=1
Theorem 9.4.3 (slide 24/ 72): slideplayer.com/slide/8024349/
I have no idea what "Theorem 9.4.3" is in your text book
but let's take it in parts
∞
Σ 1/ 5^k
k = 1
= 1/5 + 1/5^2 + 1/5^3 + ...
this is a geometric series with a = 1/5 and r = 1/5
sum∞ = a/(1-r) = (1/5)/(4/5) = 1/4
the second part is
Σ 1/ (k (k+1) ) , for k = 1 to ∞
By partial fractions i can replace 1/ (k (k+1) )
with 1/k - 1/(k+1)
so Σ 1/ (k (k+1) ) , for k = 1 to ∞
= Σ 1/k - 1/(k+1) ) , for k = 1 to n
= 1/1-1/2 + 1/2-1/3 + 1/2-1/4 + .... + 1/n - 1/(n+1_
= 1 - 1/(n+1) , notice all those other terms cancel
so when n ----> ∞ , 1/(n+1) ----> 0
thus:
Σ 1/ (k (k+1) ) , for k = 1 to ∞
= 1
and finally
∞
Σ ( (1/ 5^k) - (1/ (k (k+1) ) )
k=1
= 1/4 - 1
= -3/4
This theorem just says the summation is a linear operation. As are addition and subtraction. So, your sum is just
Σ 1/5^k - Σ 1/(k(k+1))
Σ 1/5^k is just a geometric series.
1/(k(k+1)) = 1/k - 1/(k+1) so the sum is 1 - 1/2 + 1/2 - 1/3 + ... = 1
To use Theorem 9.4.3 to find the sum of the series, we first need to write the series in the form of a telescoping series.
Given series:
∞
Σ( (1/ 5^k) - (1/ (k (k+1) ) )
k=1
To write it as a telescoping series, we need to find a common denominator and simplify the expression.
Let's start by simplifying the expression:
∞
Σ( (1/ 5^k) - (1/ (k (k+1) ) )
k=1
= ∞
Σ( (1/ 5^k) - (1/ k) + (1/ (k+1) )
k=1
Now, let's find a common denominator for the second and third terms:
= ∞
Σ( (1/ 5^k) - (k+1 - k)/ k (k+1) )
k=1
= ∞
Σ( (1/ 5^k) - 1/ k + 1/ (k+1) )
k=1
By Theorem 9.4.3, a telescoping series is defined as:
∞
Σ[ (f(k+1) - f(k) )
k=1
In our case, f(k) = 1/5^k - 1/k + 1/(k+1)
So, the sum of the given series is:
∞
Σ[ (1/5^(k+1) - 1/(k+1) + 1/(k+2)) - (1/5^k - 1/k + 1/(k+1))]
k=1
= (1/5^2 - 1/(2) + 1/(3)) - (1/5^1 - 1/(1) + 1/(2)) + (1/5^3 - 1/(3) + 1/(4)) - (1/5^2 - 1/(2) + 1/(3)) + ...
The terms in the parentheses will cancel each other out, and the sum of the series is:
1/25 - 1/5 + 1/15 = 3/25
Therefore, the sum of the series is 3/25.
To find the sum of the series ∑ ((1/5^k) - (1/(k(k+1))), we can use Theorem 9.4.3. Let's break it down step by step:
Step 1: Define the general term of the series.
The general term of the series is given by aₖ = ((1/5^k) - (1/(k(k+1))).
Step 2: Split the series into two separate series.
We can split the given series into the difference of two series:
Series A: ∑ (1/5^k)
Series B: ∑ (1/(k(k+1)))
Step 3: Find the sum of Series A.
Series A is a geometric series with a common ratio of 1/5. We can use the formula for the sum of a geometric series to find its value:
Sum of Series A = a / (1 - r), where a is the first term and r is the common ratio.
In this case, the first term a = 1/5^1 = 1/5, and the common ratio r = 1/5.
So, Sum of Series A = (1/5) / (1 - 1/5) = (1/5) / (4/5) = 1/4.
Step 4: Find the sum of Series B.
Series B is the sum of an infinite number of rational terms. To evaluate it, we can simplify it first:
1/(k(k+1)) = 1/k - 1/(k+1)
The series B can now be written as:
Series B = ∑ (1/k - 1/(k+1))
Step 5: Evaluate the telescoping sum.
By simplifying the terms, we can see that this series will telescope, meaning most of the terms will cancel out, leaving only the first and last terms.
Series B = (1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1)) + ...
Notice that in each term, the second term cancels out with the first term of the next term. This pattern continues until the last term.
So, all the intermediate terms cancel out, and we are left with the first term (1/1) and the last term (1/n+1):
Series B = 1 - 1/(n+1)
Step 6: Take the limit.
To find the sum of Series B, we take the limit as n approaches infinity:
Sum of Series B = lim(n → ∞) (1 - 1/(n+1))
Taking the limit, we get:
Sum of Series B = 1 - 0 = 1.
Step 7: Find the sum of the original series.
Now that we have the sums of Series A and Series B, we can find the sum of the original series by subtracting the sum of Series B from the sum of Series A:
Sum of the original series = Sum of Series A - Sum of Series B
= 1/4 - 1
= -3/4.
Therefore, the sum of the series ∑ ((1/5^k) - (1/(k(k+1))) is -3/4.