Question

If $S_n=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{n^2}$ and $T_n=\frac{2n-1}{n}$, then

• A. $S_2<T_2$
• B. If $S_k<T_k$ then $S_{k+1}<T_{k+1}$
• C. $S_n<T_n$ for all $n\ge 2$
• D. $S_n>T_n$ for all $n\ge 2007$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $S_2<T_2$
B If $S_k<T_k$ then $S_{k+1}<T_{k+1}$
C $S_n<T_n$ for all $n\ge 2$

$S_2=\frac{1}{1^2}+\frac{1}{2^2}=\frac{5}{4},T_2=\frac{3}{2}$
$\Rightarrow S_2<\frac{3}{2}$
Option A is true.
If $S_k<T_k$
Then $\frac{1}{1^2}+\frac{1}{2^2}+.....+\frac{1}{k^2}<2-\frac{1}{k}$
On adding $\frac{1}{(1+k{)}^2}$ on both sides $\frac{1}{1^2}+\frac{1}{2^2}+....+\frac{1}{k^2}+\frac{1}{(1+k{)}^2}<2-\frac{1}{k}+\frac{1}{(k+1{)}^2}$

$\Rightarrow S_{k+1}<2-\frac{1}{k}+\frac{1}{(k+1{)}^2}......\left(1\right)$

$T_{k+1}=2-\frac{1}{k+1}$

If we can show that the right hand side of equation (1) is less than $T_{k+1}$, then we can say $S_{k+1}<T_{k+1}$

So, we have to show that $2-(\frac{1}{k}-\frac{1}{(k+1{)}^2})<2-\frac{1}{k+1}$

which is true if $\frac{1}{k}-\frac{1}{(k+1{)}^2}>\frac{1}{k+1}$

which is true if $(k+1{)}^2-k>k(k+1)$

which is true if $k^2+1+k>k^2+k$

The above inequality is always true.

Hence, the initial inequality $2-(\frac{1}{k}-\frac{1}{(k+1{)}^2})<2-\frac{1}{k+1}$ is also true.$....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ we conclude

$S_{k+1}<T_{k+1}$

So, option B is true.

By method of mathematical induction, The statement in option B is true$\Rightarrow$ $S_n<T_n\forall n\ge 2$

Hence, option C is correct.

So option D is incorrect.