Proving $ {1 \over 2\sqrt1} + {1 \over 3\sqrt2} + \dots + {1 \over (n+1)\sqrt n} \lt 2$ using induction

Question

I have come across this problem when practicing induction. As stated in the title, it is required to prove $$ {1 \over 2\sqrt1} + {1 \over 3\sqrt2} + \dots + {1 \over (n+1)\sqrt n} \lt 2~~~ \forall n \in \Bbb{N}$$ Obviously, on its own this statement can not be proven using induction, as it's an infinite sum. It is hinted at in the study material that some term $a_n \gt 0$ should be defined for every n and added on the left so that the inequality becomes $$ {1 \over 2\sqrt1} + {1 \over 3\sqrt2} + \dots + {1 \over (n+1)\sqrt n} + a_n \le 2 $$ If we define $P(n) = \displaystyle{{1 \over 2\sqrt1} + {1 \over 3\sqrt2} + \dots + {1 \over (n+1)\sqrt n} + a_n}$, we have $$ P(n+1) = P(n) + {1 \over (n+2)\sqrt{n+1}} + a_{n+1} - a_n \le 2 $$ and if we assume $P(n) \le 2$ we get the following: $$ a_k - a_{k+1} \ge {1 \over (k+2)\sqrt{k+1}} $$ and since $ P(1) \lt 2 $, $ a_1 \le \displaystyle{\frac{3}{2}} $ must hold true. Now the problem becomes defining a sequence $a_n$ that satisfies these conditions. I have tried many and found none. Perhaps my approach itself is invalid. Please help!