In the given series there is sum of multiple of corresponding terms of two A.P.’s. The two A.P.’s are
(i)

1

1

,

1

2

,

1

3

, ... and
(ii)

1

2

,

1

3

,

1

4

, ...
Now the nth term of sum is
an = (nth term of the sequence formed by first A.P.) × (nth term of the sequence formed by second A.P.)
=

1

n

×

1

n+1

=

1

n(n+1)

=

n+1−n

n(n+1)

=

n+1

n(n+1)

−

n

n(n+1)

=

1

n

−

1

n+1

Hence, the sum to n terms is,
Sn =

n

Σ

k=1

ak =

n

Σ

k=1

(/k−

1

k+1

)
=

n

Σ

k=1

(

1

k

)−

n

Σ

k=1

(

1

k+1

) = (

1

1

+

1

2

+

1

3

+...+

1

n

) - (

1

2

+

1

3

+...+

1

n+1

)
= 1 +

1

2

−

1

2

+

1

3

−

1

3

+ ... +

1

n

−

1

n

−

1

n+1

= 1 -

1

n+1

=

n+1−1

n+1

=

n

n+1