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Chapter 9: Problem 91

Find the sum. $$\sum_{j=3}^{5} \frac{1}{j^{2}-3}$$

Short Answer

Expert verified
The sum of the values calculated based on the values of j from 3 to 5 is approximately 0.259.

Step by step solution

01

Identify the Range of j

In the given summation, j ranges from 3 to 5. For these values of j, substitute them into the given formula, \( \frac{1}{j^{2}-3} \) and compute the sum.
02

Substitute j=3

When j=3, the formula becomes \( \frac{1}{3^{2}-3} \) which simplifies to \( \frac{1}{6} \).
03

Substitute j=4

When j=4, the formula becomes \( \frac{1}{4^{2}-3} \) which simplifies to \( \frac{1}{13} \).
04

Substitute j=5

When j=5, the formula becomes \( \frac{1}{5^{2}-3} \) which simplifies to \( \frac{1}{22} \).
05

Calculate the Sum

Now sum up these fractions: \( \frac{1}{6} + \frac{1}{13} + \frac{1}{22} \). The sum comes out to be approximately 0.259.

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