How do you find the sum of #Sigma 1/(j^2-3)# where j is [3,5]?

Answer 1

# sum_(j=3)^5 = 124/429#

With so few term involved, the easiest approach is just write out the terms and compute the sum:

# sum_(j=3)^5 1/(j^2-3) = 1/(3^2-3) + 1/(4^2-3) + 1/(5^2-3)# # " " = 1/(9-3) + 1/(16-3) + 1/(25-3)# # " " = 1/(6) + 1/(13) + 1/(22)# # " " = 124/429#

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Answer 2

Scarlett Allison

To find the sum of the series ( \sum_{j=3}^{5} \frac{1}{j^2 - 3} ), you evaluate the expression for each value of ( j ) from 3 to 5 and then add them together.

Plugging in the values of ( j ), the series becomes:

[ \frac{1}{3^2 - 3} + \frac{1}{4^2 - 3} + \frac{1}{5^2 - 3} ]

[ = \frac{1}{6} + \frac{1}{13} + \frac{1}{22} ]

[ = \frac{11}{66} + \frac{5}{66} + \frac{3}{66} ]

[ = \frac{11 + 5 + 3}{66} ]

[ = \frac{19}{66} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.