How do you find the sum given #sum_(k=0)^4 1/(k^2+1)#?

Answer 1

# sum_(k=0)^(k=4) 1/(k^2+1) = 158/85#

# sum_(k=0)^(k=4) 1/(k^2+1) = 1/(0^2+1) + 1/(1^2+1) + 1/(2^2+1) + 1/(3^2+1) + 1/(4^2+1)# # :. sum_(k=0)^(k=4) 1/(k^2+1) = 1/(0+1) + 1/(1+1) + 1/(4+1) + 1/(9+1) + 1/(16+1)# # :. sum_(k=0)^(k=4) 1/(k^2+1) = 1/1 + 1/2 + 1/5 + 1/10 + 1/17# # :. sum_(k=0)^(k=4) 1/(k^2+1) = 158/85#

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Answer 2

Aurora Alvarado

To find the sum of ( \sum_{k=0}^4 \frac{1}{k^2+1} ), you evaluate the expression for each value of ( k ) from 0 to 4, and then add those values together. So, you would compute:

[ \frac{1}{0^2+1} + \frac{1}{1^2+1} + \frac{1}{2^2+1} + \frac{1}{3^2+1} + \frac{1}{4^2+1} ]

[ = \frac{1}{1} + \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17} ]

[ = 1 + \frac{1}{2} + \frac{1}{5} + \frac{1}{10} + \frac{1}{17} ]

[ = \frac{141}{85} ]

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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.