It appears that the terms of the series frac{1}{1000}+frac{1}{1001}+frac{1}{1002}+frac{1}{1003}+... are less than the corresponding terms of the conve

We have given series
$\frac{1}{1000}+\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...$
And convergent series
$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...$
Form the given two series it is clear that the first four term of the first series are less than the corresponding terms of the second series, which is convergent. But from this we can not conclude that the first series is also convergent, as we need each nth term of the first series is less than the nth term of second series.
We know Direct comparison test:
If 0 for all n
Then,
1.If $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ converges, then $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges.
2.If $\sum _{n=1}^{\mathrm{\infty }}{a}_n$ diverges, then $\sum _{n=1}^{\mathrm{\infty }}{b}_n$ diverges.
Let $a_n$ be the nth term of the first series and $b_n$ be the nth term of the second series. Then we get
$a_n=\frac{1}{999+n}$ and $b_n=\frac{1}{n^2}$
Notice that $a_n$ gives us
$\frac{1}{999+n}<\frac{1}{n^2}$
$\Rightarrow n^2-n-999<0$
$\Rightarrow (n-32.11)(n+31.11)<0$ [by quadratic formula]
It follows that $n^2-n-999>0$ for all $n\ge 33$.Therefore, 0 for all $n\le 32$ and $a_n>b_n$ for all $n\ge 33$
Therefore, given two sequence does not satisfying the direct comparison test hypothesis. So, we can not conclude anything about convergence of the first series by verifying hypothesis for the first few term of two series.
Indeed, the first series is divergent( by direct comparison test with the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$) as we know that $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ is divergent and $\frac{1}{999+n}<\frac{1}{n}$ for all $n\in \mathbb{N}$
Changing a finite number of terms in a series does not change whether or not it converges, although it may change the value of its sum if it does converge.

Answer is given below (on video)