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Thursday, April 23, 2015

Find The Sum Involving The Inverse Tangent Function

We are given to evaluate:

Sn=nk=0[tan1(1k2+k+1)]


My solution:

Using the identity:

tan1(x)=cot1(1x)

we may write:

Sn=nk=0[cot1(k2+k+1)]

Now, using the fact that:

k2+k+1=k(k+1)+1(k+1)k

and the identity:

cot(αβ)=cot(α)cot(β)+1cot(β)cot(α)

We may now write:

Sn=nk=0[cot1(k)cot1(k+1)]

This is a telescoping series, hence:

Sn=cot1(0)cot1(n+1)=π2cot1(n+1)

Using the identity:

tan1(x)+cot1(x)=π2

We may also write:

Sn=tan1(n+1)

And so we have found:

nk=0[tan1(1k2+k+1)]=tan1(n+1)

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