Using the principle of mathematical induction , prove that for `n in N , (1)/(n+1) + (1)/(n+2) + (1)/(n+3) + "……." + (1)/(3n+1) gt 1`.

Explanation

`P(n): (1)/(n+1)+(1)/(n+2)+(1)/(n+3)+"....."+(1)/(3n+1) gt 1`, fpr all ` n in N`
For `n = 1`,
`L.H.S. = 1/2 + 1/3 + 1/4 = 13/12 gt 1`
Thus, P(1) is true.
Let `P(n)` be true for some `n = k`.
i.e., `(1)/(k+1)+(1)/(k+2)+(1)/(k+3) +"......."+(1)/(3k+1) gt 1"........"(1)`
Adding `(1)/(3k+4)` on both sides, we get
`(1)/(k+1)+(1)/(k+2)+(1)/(k+3) +"......."+(1)/(3k+1) + (1)/(3+4) gt 1 + (1)/(3k+4)`
Clearly `1+(1)/(3k+4) gt 1`
so, `(1)/(k+1)+(1)/(k+2)+(1)/(k+3) +"......."+(1)/(3k+1) +(1)/(3k+4) gt 1`
`or `(1)/(k+1)+(1)/(k+2)+(1)/(k+3) +"......."+(1)/(3(k+1) 1+) gt 1`
Thus, `P(k+1)` is true whenever `P(k)` is true.
Hence, by the principle of mathemetical induction statement `P(n)` is true for all natural numbers.