Kvant Math Problem 812
The sum is
Verified: yes
Verdicts: PASS + PASS
Solve time: 13m31s
Source on kvant.digital
Problem
Prove that for every positive integer $n$, $$\dfrac12+\dfrac1{3\sqrt{2}}+\dfrac1{4\sqrt{3}}+\ldots+\dfrac1{(n+1)\sqrt{n}}\lt2.$$
S. I. Maizus
Exploration
The sum is
$$S_n=\frac12+\frac1{3\sqrt2}+\frac1{4\sqrt3}+\cdots+\frac1{(n+1)\sqrt n} =\sum_{k=1}^{n}\frac1{(k+1)\sqrt k}.$$
A natural idea is to compare each term with a telescoping difference involving $\sqrt{k}$.
For $k=1,2,3$,
$$\frac1{(k+1)\sqrt k} = \frac12,\ \frac1{3\sqrt2},\ \frac1{4\sqrt3},$$
while
$$\frac2{\sqrt k+\sqrt{k+1}} = \frac2{1+\sqrt2},\ \frac2{\sqrt2+\sqrt3},\ \frac2{\sqrt3+2}.$$
Numerically,
$$\frac12<\frac2{1+\sqrt2},\qquad \frac1{3\sqrt2}<\frac2{\sqrt2+\sqrt3},\qquad \frac1{4\sqrt3}<\frac2{\sqrt3+2}.$$
Since
$$\frac2{\sqrt k+\sqrt{k+1}} =\sqrt{k+1}-\sqrt k,$$
this comparison would yield a telescoping sum.
The crucial point is to prove
$$\frac1{(k+1)\sqrt k} < \sqrt{k+1}-\sqrt k = \frac1{\sqrt{k+1}+\sqrt k} \cdot 2.$$
After clearing denominators, this becomes
$$\sqrt{k+1}+\sqrt k<2(k+1)\sqrt k,$$
which is immediate because $\sqrt{k+1}<\sqrt k+1$, hence the left-hand side is less than $2\sqrt k+1$, whereas the right-hand side equals $2k\sqrt k+2\sqrt k$ and exceeds $2\sqrt k+1$ for every positive integer $k$.
The resulting telescoping sum gives
$$S_n<2(\sqrt{n+1}-1).$$
This alone is not enough for large $n$. A sharper comparison is needed.
Trying instead
$$\frac1{(k+1)\sqrt k} < \frac2{\sqrt k}-\frac2{\sqrt{k+1}} = \frac{2(\sqrt{k+1}-\sqrt k)}{\sqrt k,\sqrt{k+1}},$$
we obtain, after simplification,
$$\frac1{k+1} < \frac{2}{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)}.$$
Multiplying by positive quantities,
$$\sqrt{k+1}+\sqrt k<2\sqrt{k+1},$$
which is true because $\sqrt k<\sqrt{k+1}$.
This gives a perfect telescoping estimate:
$$S_n < 2\sum_{k=1}^{n}\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}!}\right) = 2\left(1-\frac1{\sqrt{n+1}}\right) <2.$$
The delicate step is the termwise inequality leading to the telescoping expression.
Problem Understanding
We must prove that for every positive integer $n$,
$$\sum_{k=1}^{n}\frac1{(k+1)\sqrt k}<2.$$
This is a Type B problem, a pure proof.
The core difficulty is to find a suitable upper bound for the general term that produces a telescoping series. Once such a bound is obtained, the sum collapses to a simple expression that is visibly less than $2$.
Proof Architecture
The proof uses one lemma.
Lemma. For every positive integer $k$,
$$\frac1{(k+1)\sqrt k} < 2\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right).$$
Sketch. Rewrite the right-hand side as
$$\frac{2(\sqrt{k+1}-\sqrt k)}{\sqrt k,\sqrt{k+1}} = \frac{2}{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)},$$
and compare denominators.
After summing the lemma from $k=1$ to $n$, the right-hand side telescopes, yielding
$$S_n < 2\left(1-\frac1{\sqrt{n+1}}\right) <2.$$
The lemma is the step most likely to fail under scrutiny, because the entire argument depends on obtaining exactly the correct telescoping bound.
Solution
Let
$$S_n=\sum_{k=1}^{n}\frac1{(k+1)\sqrt k}.$$
For every positive integer $k$,
$$2\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right) = \frac{2(\sqrt{k+1}-\sqrt k)}{\sqrt k,\sqrt{k+1}}.$$
Using
$$\sqrt{k+1}-\sqrt k = \frac1{\sqrt{k+1}+\sqrt k},$$
we obtain
$$2\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right) = \frac{2}{\sqrt{k+1},(\sqrt{k+1}+\sqrt k)}.$$
Hence
$$2\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right) - \frac1{(k+1)\sqrt k} = \frac{2}{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)} - \frac1{(k+1)\sqrt k}.$$
Since all quantities are positive, it suffices to compare denominators. The inequality
$$\sqrt k<\sqrt{k+1}$$
implies
$$\sqrt{k+1}+\sqrt k<2\sqrt{k+1}.$$
Multiplying by $\sqrt{k+1}$ gives
$$\sqrt{k+1},(\sqrt{k+1}+\sqrt k) < 2(k+1).$$
Multiplying by $\sqrt k$ yields
$$\sqrt{k+1},(\sqrt{k+1}+\sqrt k)\sqrt k < 2(k+1)\sqrt k.$$
Taking reciprocals,
$$\frac{2}{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)} > \frac1{(k+1)\sqrt k}.$$
Therefore
$$\frac1{(k+1)\sqrt k} < 2\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right).$$
Summing this inequality for $k=1,2,\dots,n$ gives
$$S_n < 2\sum_{k=1}^{n} \left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right).$$
The series on the right telescopes:
$$\sum_{k=1}^{n} \left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right) = 1-\frac1{\sqrt{n+1}}.$$
Consequently,
$$S_n < 2\left(1-\frac1{\sqrt{n+1}}\right) <2.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the identity
$$\sqrt{k+1}-\sqrt k = \frac1{\sqrt{k+1}+\sqrt k}.$$
Multiplying the right-hand side by $\sqrt{k+1}+\sqrt k$ gives $1$, while
$$(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k) = (k+1)-k = 1,$$
so the identity is exact.
The second delicate step is the comparison of denominators. Starting from
$$\sqrt k<\sqrt{k+1},$$
we obtain
$$\sqrt{k+1}+\sqrt k<2\sqrt{k+1}.$$
Multiplication by the positive number $\sqrt{k+1}\sqrt k$ preserves the inequality and yields
$$\sqrt{k+1}(\sqrt{k+1}+\sqrt k)\sqrt k < 2(k+1)\sqrt k.$$
Since both sides are positive, reciprocation reverses the inequality, producing precisely the desired termwise estimate.
A common error would be to telescope with
$$\sqrt{k+1}-\sqrt k$$
instead of
$$\frac1{\sqrt k}-\frac1{\sqrt{k+1}}.$$
The former leads to an upper bound growing like $\sqrt n$, which is insufficient for proving a uniform bound by $2$.
Alternative Approaches
A closely related proof begins with
$$\frac1{(k+1)\sqrt k} < \frac1{k\sqrt k} = \frac1{k^{3/2}}.$$
One then proves
$$\frac1{k^{3/2}} < 2\left(\frac1{\sqrt k}-\frac1{\sqrt{k+1}}\right),$$
which follows from the same computation used in the main solution. Summing again yields
$$S_n < 2\left(1-\frac1{\sqrt{n+1}}\right) <2.$$
The main approach is preferable because it compares the original term directly with the telescoping difference, avoiding an intermediate estimate and making the source of the telescoping structure transparent.