Prove that for , we have

and further that,

We prove each of these separately (and we use the first one in the proof of the second).

*Proof #1.*We prove each of the inequalities separately. First, we prove

This just requires some algebraic manipulation,

Since for all , we have the left inequality. For the right inequality we proceed similarly,

Thus, we have the inequality on the right true for all as well

* Proof #2. * Now, we want to use the result in the first proof to prove the second set of inequalities.

First, we consider the left inequality. For the case we have

on the left and right respectively. The inequality then holds since,

Thus, the inequality holds in the case . Assume then that the inequality holds for some . Then,

But, using the first part again we know ; hence,

Thus, the inequality is true for all .

Now, for the inequality on the right. For the case we have

on the left and right, respectively. But then, since we have,

Hence, the inequality is true for the case . Assume then that it is true for some . Then,

Hence the right inequality holds for all .

Therefore we have established both halves of the inequality for all

Show that the set A given by

A := n

2021 · n +

1

n2

: n ∈ N

o

is unbounded above.

I think that this proof is supposed to be about the telescoping property

rather than induction.

To prove:

You just need to sum the relevant inequality proved in part 1 and use

the telescoping property:

and apply .

The other inequality is slightly more complicated because

and we want . The clue is that that is

valid for all wheras is only valid for . Hence we look at:

The sum is only valid for as we expected. We can then add 1

to either side of the inequlity to find:

You can also get the left inequality of part 2 by telescoping

in proof #2 , left inequality , the last step you make ,when you use the first part again ,you change small value for a bigger one but you dont know if the new left side is still going to be less than the right side.

i,too,found the same error.

tambem reparei isso

Fist of all great work that you are doing here!

In proof #2 there is a typo in the 1st line on the right side of the equation: a ‘2’ in the square root instead of a ‘k’.

Everything that follows is stated false. Of course, fixing the typo makes the proof much easier ;)