- #26

Office_Shredder

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Looks good to me now!

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- Thread starter arhzz
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- #26

Office_Shredder

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Looks good to me now!

- #27

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Excellent,thank you for your help,All of you!

- #28

epenguin

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? Can we just write

$$\left( 1+k\right) ^{1/2}-k^{1/2}=\left[ k\left( 1+\dfrac {1}{k}\right) \right] ^{1/2}-k^{1/2}$$

$$=k^{1/2}\left[ \left( 1+\dfrac {1}{k}\right) ^{1/2}-1\right] $$

For large ##k## the quantity in round brackets is approximately equal to, and more importantly, larger than ##\left( 1+\dfrac {1}{2k}\right) ## . Thus the whole for large k is approximately equal to but greater than ##\dfrac {1}{2k^{1/2}}## . The expression is reduced to a single term.

You perhaps recognise the infinite series of such terms as divergent, or else its divergence can be proved by the integral test or otherwise.

$$\left( 1+k\right) ^{1/2}-k^{1/2}=\left[ k\left( 1+\dfrac {1}{k}\right) \right] ^{1/2}-k^{1/2}$$

$$=k^{1/2}\left[ \left( 1+\dfrac {1}{k}\right) ^{1/2}-1\right] $$

For large ##k## the quantity in round brackets is approximately equal to, and more importantly, larger than ##\left( 1+\dfrac {1}{2k}\right) ## . Thus the whole for large k is approximately equal to but greater than ##\dfrac {1}{2k^{1/2}}## . The expression is reduced to a single term.

You perhaps recognise the infinite series of such terms as divergent, or else its divergence can be proved by the integral test or otherwise.

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- #29

- 147

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That is definetly a legit way to do it,although it goes into that teritory I am not comftorable with yet,where you have to make a lot of good educated guesses.Still thank you for the insight!? Can we just write

$$\left( 1+k\right) ^{1/2}-k^{1/2}=\left[ k\left( 1+\dfrac {1}{k}\right) \right] ^{1/2}-k^{1/2}$$

$$=k^{1/2}\left[ \left( 1+\dfrac {1}{k}\right) ^{1/2}-1\right] $$

For large ##k## the quantity in round brackets is approximately equal to, and more importantly, larger than ##\left( 1+\dfrac {1}{2k}\right) ## . Thus the whole for large k is approximately equal to but greater than ##\dfrac {1}{2k^{1/2}}## . The expression is reduced to a single term.

You perhaps recognise the infinite series of such terms as divergent, or else its divergence can be proved by the integral test or otherwise.

- #30

epenguin

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You will probably later come across

**Always connect!**

It is what you are using in the standard school arithmetical calculation of square roots!

And then again do you notice the the similarity between ##\left( 1+k\right) ^{1/2}-k^{1/2}## and the basic formula for all differentiation ##[ f(x + δ) - f(x)]/δ## ? Indeed the first formula*is *the second for ##f(k)= √k## and ##δ = 1 ##. Now we usually posit that ##δ ## be small - that is only in comparison with ##f(k)## so if we let ##k## be large enough, 1 is a small by comparison and our formula is a good approximation to the derivative; as we let ##k## increase to infinity it *becomes *the derivative.

$$\sqrt {1+k}-\sqrt {k}=\dfrac {d\sqrt {k}}{dk}$$

What about the sum of the series? We can say approximately that for some large ##k## called ##K##

$$\sum ^{\infty }_{k=K}(\sqrt {1+k}-\sqrt {k})= \sum ^{\infty }_{k=K}\dfrac {d\sqrt {k}}{dk}$$

and we can approximate the sum by an integral, which is of the easiest

$$\int ^{\infty }_{K}\dfrac {d\sqrt {k}}{dk}.dk=\left[ \sqrt {k}\right] ^{\infty }_{K}$$

as before and this is infinite.

No doubt this needs tidying up for more rigorous expression but I think a version of that stands up.

I don't do that now as I see something better for the problem, but we and other contributions have not totally wasted our time. Still, you practically had this yourself before:

$$\sum ^{N}_{k=1}(\sqrt {1+k}-\sqrt {k}) = $$

## √2 - √1 + √3 - √2 + √4 - √3 +... + √(N+1) - √N##

## = √(N + 1) - 1##

which increases with ##N## without limit.

You will probably some few times come across the approximation ## √(1 + a) ≈ (1 + a/2)## for small ##a## in treatments in various physics and chemistry. P But further!That is definetly a legit way to do it,although it goes into that teritory I am not comftorable with yet,where you have to make a lot of good educated guesses.Still thank you for the insight!

It is what you are using in the standard school arithmetical calculation of square roots!

And then again do you notice the the similarity between ##\left( 1+k\right) ^{1/2}-k^{1/2}## and the basic formula for all differentiation ##[ f(x + δ) - f(x)]/δ## ? Indeed the first formula

$$\sqrt {1+k}-\sqrt {k}=\dfrac {d\sqrt {k}}{dk}$$

What about the sum of the series? We can say approximately that for some large ##k## called ##K##

$$\sum ^{\infty }_{k=K}(\sqrt {1+k}-\sqrt {k})= \sum ^{\infty }_{k=K}\dfrac {d\sqrt {k}}{dk}$$

and we can approximate the sum by an integral, which is of the easiest

$$\int ^{\infty }_{K}\dfrac {d\sqrt {k}}{dk}.dk=\left[ \sqrt {k}\right] ^{\infty }_{K}$$

as before and this is infinite.

No doubt this needs tidying up for more rigorous expression but I think a version of that stands up.

I don't do that now as I see something better for the problem, but we and other contributions have not totally wasted our time. Still, you practically had this yourself before:

$$\sum ^{N}_{k=1}(\sqrt {1+k}-\sqrt {k}) = $$

## √2 - √1 + √3 - √2 + √4 - √3 +... + √(N+1) - √N##

## = √(N + 1) - 1##

which increases with ##N## without limit.

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