Am I allowed to solve this series this way?

  • 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.
 
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  • #29
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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.
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!
 
  • #30
epenguin
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You will probably later come across
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!
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! 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.
 
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