Looking for help to understand the Integral Test

[opus]

Please see the attached images that reference the text.
To my understanding, we wish to use the integral test to compare to a series to see if the series converges or diverges.
In these two examples, we use $\sum_{n=1}^\infty \frac{1}{n}$ compared with $\int_1^{k+1} \frac{1}{x}dx$ and for the second example we have $\sum_{n=1}^\infty \frac{1}{n^2}$ compared with $\int_1^k \frac{1}{x^2}dx$

Now I have a couple uncertainties about this. First, the difference in the "k+1" upper bound for the first example and the "k" upper bound for the second example. Why are they different? And this leads to my second question. In the first picture of the first example compared to the second picture of the second example, we are starting at different places on the graphs. The second example has an extra "block" to the series, and we are adding a +1 to the integral of the second example. Additionally, in the second example, the graph is above the series whereas the graph is below the series for the first example.

I understand that we are in a way trying to "bound" the series, but I don't know what's going on with the differences between these two.

Any help would be greatly appreciated. Thank you.

 

[scottdave]

Sliding the blocks over by 1 position either direction is the same as adding or subtracting a single finite block.
If you have something that converges, then adding or subtracting some finite amount will not change that fact.

Similarly, if it diverges, then adding or subtracting a finite amount will not change that fact.

Think of it like this: if the integral converges, then we can arrange a series such that it fits inside the integral.
Since the area in all the blocks is less than the area of the integral, then the series must also converge.
Since there are gaps, we still do not know the value of the series, just that it is less than the integral.
Note that if the integral diverges, then showing the blocks are less than the integral proves nothing.

Similarly, you can arrange blocks so that the area of the blocks is clearly greater than the area of the integral.
If the integral diverges, then you know the blocks are greater than that, so the series also diverges.
If the integral converges, but you arrange the blocks to be bigger than the integral, you have proved nothing.

 

[opus]

So really it's just rearranging the blocks so that we can compare them to something that we know something about. And the shift really doesn't matter because in the long run, a single block or two isn't going to make any difference at all?

 

[scottdave]

So really it's just rearranging the blocks so that we can compare them to something that we know something about. And the shift really doesn't matter because in the long run, a single block or two isn't going to make any difference at all?

Yes, you got it.

 

[opus]

Awesome, thanks scottdave!