hi guys! i need help in solving this induction problem!!
I try solving it! but i don't know if is correct very confusing question!
My solution
Your base case is incorrectly worked out to start with.
The sum from i=1 to i=2 is 1/3+1/7=10/21
which is <1 so the base case is ok.
You have the P(k) proposition as
$\displaystyle \sum_{i=1}^k\left(\frac{1}{i^2+i+1}\right)\le\ 2-\frac{1}{k-1}$
Then your P(k+1) proposition is
$\displaystyle \sum_{i=1}^{k+1}\left(\frac{1}{i^2+i+1}\right)\le\ 2-\frac{1}{k+1-1}$
Then you try to prove that IF P(k) is true THEN P(k+1) will be true.
Do this by using the fact that if Pk) is true,
P(k+1) LHS will be less than or equal to
$\displaystyle 2-\frac{1}{k-1}+\frac{1}{(k+1)^2+(k+1)+1}$
and hopefully this will be $\displaystyle \le\ 2-\frac{1}{k}$
Therefore we can ask if
$\displaystyle \frac{1}{k^2+2k+1+k+2}\le\ \frac{1}{k-1}-\frac{1}{k}\;\;?$
$\displaystyle \frac{1}{k^2+3k+3}\le\frac{k-(k-1)}{(k-1)k}\;\;?$
$\displaystyle \frac{1}{k^2+3k+3}\le\frac{1}{k^2-k}\;\;?$
$\displaystyle k^2+3k+3\ge\ k^2-k\;\;?$
$\displaystyle 4k+3\ge\ 0\;\;?$
Didn't you add the part in yellow (at the top of your third picture) to both sides in your effort?
If the Left is less than or equal to the Right in that case initially,
it will be afterwards, since you only added the same thing to both sides,
so it's pointless.
You must understand what you are trying to do with the "k" and "k+1" concept.
I've shown you most of the steps.
Study it and try to answer the final line