The comparison test for series

## The comparison test for series
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            ## Telescoping series
            In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.
            
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            ## Telescoping series
            In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.

Or can we?
            
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            ## Telescoping series
            In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.

Or can we?

Not if $p\notin\mathbb N$.
            For example, try to make the telescoping sum argument for $p=\frac32$.
            
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            ## Telescoping series
            In the same way, we can show that, for any $p>0$, the series $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+p)}$ converges, and calculate its limit.

Or can we?

Not if $p\notin\mathbb N$.
            For example, try to make the telescoping sum argument for $p=\frac32$.

You never get the cancellation, because the terms are "missing" each other.

Let's try another approach.
            
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            ## Comparison of series
            We know $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+1)}=\sum\_{k=1}^\infty a_k$ converges.
            Try $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)} = \sum\_{k=1}^\infty b_k$.

Here $0 \leqslant b_k \leqslant a_k$ so we are summing smaller things.

So $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)}$ must also converge.
            And we could replace $2$ with any real $p>1$.
            
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            ## Comparison of series
            We know $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+1)}=\sum\_{k=1}^\infty a_k$ converges.
            Try $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)} = \sum\_{k=1}^\infty b_k$.

Here $0 \leqslant b_k \leqslant a_k$ so we are summing smaller things.

So $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)}$ must also converge.
            And we could replace $2$ with any real $p>1$.

#### Theorem (Comparison test)
            If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.
            
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            ## Comparison of series
            We know $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+1)}=\sum\_{k=1}^\infty a_k$ converges.
            Try $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)} = \sum\_{k=1}^\infty b_k$.

Here $0 \leqslant b_k \leqslant a_k$ so we are summing smaller things.

So $\displaystyle\sum\_{k=1}^\infty \frac1{k(k+2)}$ must also converge.
            And we could replace $2$ with any real $p>1$.

#### Corollary
            If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty b_k$ diverges, then so too does $\displaystyle\sum\_{k=1}^\infty a_k$.
            
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            ## Comparison test
            #### Theorem (Comparison test)
            If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.

#### Example
            $\sum\_{k=1}^\infty \frac1{k^2}$ converges.

#### Idea
            Want to use comparison with $\sum\_{k=1}^\infty \frac1{k(k+1)}$.
            But can't because $0 \leqslant \frac1{k^2} \leqslant \frac1{k(k+1)}$ is false!

#### Proof
            $\sum\_{k=1}^\infty \frac1{k^2} = \frac11 + \sum\_{k=2}^\infty \frac1{k^2} = \frac11 + \sum\_{k=1}^\infty \frac1{(k+1)^2}$.
            But $0 \leqslant \frac1{(k+1)^2} \leqslant \frac1{k(k+1)}$ and $\sum\_{k=1}^\infty \frac1{k(k+1)}$ converges.
            Therefore, by the comparison test, so too does $\sum\_{k=1}^\infty \frac1{(k+1)^2}$.

Hence $\sum\_{k=1}^\infty \frac1{k^2}$ converges.   □
            
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            ## Comparison test
            #### Theorem (Comparison test)
            If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.

#### Example
            $\displaystyle \sum\_{k=1}^\infty \frac1{k^t}$ converges for all $t\geq2$ and diverges for all $t\leq1$.
            
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            ## Comparison test
            #### Theorem (Comparison test)
            If, for all $k\in\mathbb N$, $0 \leqslant b_k \leqslant a_k$, and $\displaystyle\sum\_{k=1}^\infty a_k$ converges, then so too does $\displaystyle\sum\_{k=1}^\infty b_k$.

#### Example
            $\displaystyle \sum\_{k=1}^\infty \frac1{k^t}$ converges for all $t\geq2$ and diverges for all $t\leq1$.

This follows by comparison with $\displaystyle\sum\_{k=1}^\infty \frac1{k^2}$ and $\displaystyle\sum\_{k=1}^\infty \frac1k$.

We still don't know about $1<t<2$.
            See later.