# How do you find the sum of #Sigma 1/(k^2+1)# where k is [0,3]?

# sum_(k=0)^(3) 1/(k^2+1) = 9 /5 #

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To find the sum of ( \Sigma \frac{1}{k^2+1} ) where ( k ) ranges from 0 to 3, you simply plug in the values of ( k ) from 0 to 3 into the expression and then add up the results.

( k = 0: \frac{1}{0^2 + 1} = \frac{1}{1} = 1 )

( k = 1: \frac{1}{1^2 + 1} = \frac{1}{2} )

( k = 2: \frac{1}{2^2 + 1} = \frac{1}{5} )

( k = 3: \frac{1}{3^2 + 1} = \frac{1}{10} )

Summing up these results:

( 1 + \frac{1}{2} + \frac{1}{5} + \frac{1}{10} = \frac{26}{10} = \frac{13}{5} )

So, the sum of ( \Sigma \frac{1}{k^2+1} ) where ( k ) ranges from 0 to 3 is ( \frac{13}{5} ).

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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.

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.

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.

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