I am trying to solve a sigma/summation notation problem:
n = 99, i = 4 and each term in the sequence is determined by (1/i) - (1/(i+1)).
Since n = 99, I think it's safe to assume my professor does not want me to actually go through the process of subbing in "i" 99 times to find the answer.
How should I approach this question? Do I evaluate this question as a limit approaching 99? Also does the fact that the "i" is in the denominator of a rational create any extra steps I need to take? Thank you for the help.
Note that the series is collapsing. All the terms except the first and the last cancel out!
(1/4 - 1/5) + (1/5 - 1/6) ... + (1/99 - 1/100) = 1/4 - 1/100
Ah! Oh my, I completely missed that! Thank you for the help Steve.
Always be on the lookout for trick questions...
Develop your inner laziness!
To solve the summation notation problem, you can use the telescoping property. This property allows you to simplify the summation by canceling out terms that appear both as the numerator and denominator in consecutive terms.
Based on the given sequence, each term is (1/i) - (1/(i+1)). Let's analyze this expression:
1/i - 1/(i+1) = (i+1)/(i(i+1)) - i/(i(i+1))
= (i+1 - i)/(i(i+1))
= 1/(i(i+1))
Now, we can rewrite the summation notation using this simplified expression:
∑(1/i - 1/(i+1)) from i = 4 to n
In this case, i starts from 4 and goes up to n = 99. We can limit the terms in the sum by using the telescoping property. Let's see how:
∑(1/i - 1/(i+1)) from i = 4 to n
= (1/4 - 1/5) + (1/5 - 1/6) + … + (1/98 - 1/99) + (1/99 - 1/100)
As we can observe, each term cancels out with the subsequent term, except for the first term (1/4) and the last term (1/99). Therefore, we are left with:
(1/4) - (1/99)
Now, you can evaluate this expression to calculate the sum without having to substitute i 99 times. You can simplify it further, if needed.
Note that evaluating the sum as a limit approaching 99 would not be necessary in this case, as the telescoping property allows you to simplify the summation without performing each individual calculation.
Hope this explanation helps you solve the problem more efficiently.