 Hello and welcome to the session. In this session we discuss the following question which says find the nth term, sum of nth terms and sum to infinity of the series 1 upon 1 into 2 plus 1 upon 2 into 3 plus 1 upon 3 into 4 plus and so on. Let's now move on to the solution. Now the given series is 1 upon 1 into 2 plus 1 upon 2 into 3 plus 1 upon 3 into 4 plus and so on. And we are supposed to find the nth term, sum of the n terms and sum to infinity of this given series. Now the nth term of the series is given by Tn and this would be equal to 1 upon nth term of 1, 2, 3 and so on. This into nth term of 2, 3, 4 and so on. Now the nth term of 1, 2, 3 and so on would be n and nth term of 2, 3, 4 and so on would be n plus 1. So here we have Tn is equal to 1 upon n into n plus 1 the whole. Or we can write this numerator 1 as n plus 1 minus n and this whole upon n into n plus 1 the whole. So this is equal to n plus 1 upon n into n plus 1 the whole minus n upon n into n plus 1 the whole. This n plus 1, n plus 1 cancels n and n cancels and so this is equal to 1 upon n minus 1 upon n plus 1. This is the nth term of the given series. Now next putting n equal to 1, 2, 3 and so on up to n we get T1 is equal to 1 minus 1 upon 1 plus 1 which is 2 then T2 is equal to 1 upon 2 minus 1 upon 2 plus 1 which is 3 then T3 is equal to 1 upon 3 minus 1 upon 3 plus 1 which is 4 and so on up to Tn which is equal to 1 upon n minus 1 upon n plus 1. Now to find the sum of n terms of the given series which would be given by Sn and this would be equal to sum of these terms of the series that is T1 plus T2 plus T3 plus and so on up to Tn. Now substituting the values for T1, T2, T3 and so on up to Tn we get this is equal to now T1 is 1 minus 1 upon 2 so 1 minus 1 upon 2 plus T2 that is 1 upon 2 minus 1 upon 3 plus T3 that is 1 upon 3 minus 1 upon 4 plus and so on plus 1 upon n minus 1 minus 1 upon n that is this is the n minus 1 s term of the series plus the n h term of the series which is 1 upon n minus 1 upon n plus 1. Now all these terms would get cancelled we are left with 1 minus 1 upon n plus 1 that is Sn which is the sum of the n terms of the series is equal to 1 minus 1 upon n plus 1 or this is equal to n plus 1 minus 1 upon n plus 1 1 minus 1 cancels and this is equal to n upon n plus 1 or you can say that Sn is equal to 1 upon 1 plus 1 upon n so this is the sum of the n terms of the series now sum to infinity of the series would be equal to S infinity and this is equal to limit n tends to infinity Sn or you can say limit n tends to infinity 1 upon 1 plus 1 upon n now we know that as n tends to infinity 1 upon n tends to 0 therefore S infinity is equal to 1 upon 1 plus 0 which is equal to 1 that is S infinity is equal to 1 and this is the sum to infinity of the series. So finally we have the l h term of the series is equal to 1 upon n minus 1 upon n plus 1 then sum to n terms of the series that is Sn is equal to 1 upon 1 plus 1 upon n so this is 1 upon 1 plus 1 upon n and sum to infinity of the series that is S infinity is equal to 1 so this is the final answer of the question so this completes the session hope you have understood the solution of this question.