Question 4

Find the sum to n terms of the series fraction numerator 1 over denominator 1 cross times 2 end fraction space plus fraction numerator 1 over denominator 2 cross times 3 end fraction space plus space fraction numerator 1 over denominator 3 space cross times space 4 end fraction space plus space...

Answer

The given series is fraction numerator 1 over denominator 1 cross times 2 end fraction space plus space fraction numerator 1 over denominator 2 cross times 3 end fraction space plus fraction numerator 1 over denominator 3 cross times 4 end fraction space plus space...

nth term, an = fraction numerator 1 over denominator n open parentheses n plus 1 close parentheses end fraction equals 1 over n minus fraction numerator 1 over denominator n plus 1 end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis B y space p a r t i a l space f r a c t i o n s right parenthesis

a subscript 1 space equals space 1 over 1 space minus space 1 half
a subscript 2 space equals space 1 half space minus space 1 third
a subscript 3 space space equals space 1 third space minus space 1 fourth...
a subscript n space equals space 1 over n space minus space fraction numerator 1 over denominator n plus 1 end fraction

Adding the above terms column wise, we obtain

a subscript 1 space plus space a subscript 2 space plus space... space plus space a subscript n space equals space open square brackets 1 over 1 plus 1 half plus 1 third plus...1 over n close square brackets minus open square brackets 1 half plus 1 third plus 1 fourth plus... fraction numerator 1 over denominator n plus 1 end fraction close square brackets
therefore space S subscript n space equals space 1 minus space fraction numerator 1 over denominator n plus 1 end fraction equals fraction numerator n plus 1 minus 1 over denominator n plus 1 end fraction equals fraction numerator n over denominator n plus 1 end fraction

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