My Math Forum Algebra trick to make this induction proof work?

 Algebra Pre-Algebra and Basic Algebra Math Forum

 August 25th, 2013, 03:08 AM #1 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Algebra trick to make this induction proof work? Hello there, For two days, I have tried to prove that this is true using induction: $\text{Pn: } \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} \,\>\, 2(\sqrt{n+1}-1)$ $P_1$ is true. I checked. The best I could ever arrive at, assuming $P_k$ to be true, was: $1 \, > \, 2(\sqrt{k+2}\sqrt{k+1}) - 2k + 2$. However, I suspect there is a prettier solution, and I am unable to continue my happy life without knowing what it was. Are any of you able to work it out? Thank you for your time. Kind regards, Marius
 August 25th, 2013, 05:45 AM #2 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: Algebra trick to make this induction proof work? I found the solution myself: $P_{k+1}:$ $2\sqrt{k+1} + \frac{1}{\sqrt{k+1}} \,>\,2\sqrt{k+2}$ $\frac{2\sqrt{k+1}^2+1}{\sqrt{k+1}} \,>\,2\sqrt{k+2}$ $\frac{2k+2+1}{\sqrt{k+1}} \,>\,2\sqrt{k+2}$ $(\frac{2k+3}{\sqrt{k+1}})^2 \,>\,(2\sqrt{k+2})^2$ $\frac{(2k+3)^2}{k+1} \,>\,4(k+2)$ $(2k+3)^2 \,>\,4(k+2)(k+1)$ $9 \,>\, 8$ Thank you for your time. Kind regards, Marius
 August 25th, 2013, 05:52 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 Your work is okay. Obviously, ?(k+2)/?(k+1) + 1 > 2. Multiplying by ?(k+2)/?(k+1) - 1 gives 1/(k+1) > 2(?k+2)/?(k+1) - 1), which implies 1 > 2(?(k+2)?k+1)) - 2(k + 1), as needed to complete the induction proof.
 August 25th, 2013, 05:56 AM #4 Senior Member   Joined: Sep 2010 From: Oslo, Norway Posts: 162 Thanks: 2 Re: Algebra trick to make this induction proof work? Thanks skipjack, what do you think of my other attempt? Thanks.M
 August 25th, 2013, 07:08 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,972 Thanks: 2222 Both are okay. The first is abbreviated and unfinished, whilst the second fills out the first and finishes it.

 Tags algebra, induction, make, proof, trick, work

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post fe phi fo Calculus 4 June 12th, 2012 08:47 AM DenJansen Real Analysis 1 February 20th, 2012 09:33 PM restin84 Applied Math 3 October 10th, 2009 07:22 AM mikeportnoy Algebra 3 February 18th, 2009 11:51 AM MaD_GirL Number Theory 5 November 14th, 2007 06:34 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top