20100128, 06:52  #1 
Sep 2004
13·41 Posts 
can anyone here do algebra :)
apparently I can't lol. I'm trying to do a proof by induction and I think I'm getting tripped up on the algebra.
so prove some of harmonic series is = (n+1) * harmonic nth term  n so 1 = 2*11 now for tricky part (n+1) Hn  n + Hn+1 = (n+2)Hn+1  (n+1) where Hn is nth harmonic term so I think this reduces to (n+1)/n  n + 1/(n+1) = (n+2)/(n+1)  (n+1) 1 + 1/n  n + 1/(n+1) = (n+2)/(n+1)  n  1 (2n+1) / (n*(n+1)) + 1 =... I tried going farther but didn't get anywhere... Thanks guys! 
20100128, 09:35  #2 
(loop (#_fork))
Feb 2006
Cambridge, England
2·29·109 Posts 
You're not writing down clearly what you're trying to prove, which is always a bad start.
Let A = 1+1/2+1/3+1/4 = 25/12 Let B = 1+1/2+1/3+1/4+1/5 = 137/60 and you seem to be wanting to prove that B = 5A4, which is not the case. 
20100128, 09:47  #3  
Sep 2004
13×41 Posts 
Quote:
I'm sorry I can't state it any better. Last fiddled with by Joshua2 on 20100128 at 09:50 

20100128, 10:29  #4  
"William"
May 2003
New Haven
2^{2}·3^{2}·5·13 Posts 
Quote:
(1+1/2+1/3+1/4+1/5)=6*(1/4)5 or, depending on the meaning of "harmonic nth term" (1+1/2+1/3+1/4+1/5)=6*(1/5)5 or possibly (1+1/2+1/3+1/4+1/5)=6*(25/12)5 Which one is it? 

20100128, 22:10  #5 
Sep 2004
1000010101_{2} Posts 
Thanks guys, it appears none are true plugging into a calculator. So I made a mistake somewhere before? I found out that H sub 4 is 25/12
Last fiddled with by Joshua2 on 20100128 at 22:15 
20100128, 22:13  #6 
Sep 2004
1025_{8} Posts 
Here is the picture of the problem...

20100128, 23:36  #7 
Jun 2003
5×23×41 Posts 
This is correct. Here Hn is _not_ the harmonic nth term, it is the sum of the harmonic series upto the nth term.

20100128, 23:42  #8 
"William"
May 2003
New Haven
2340_{10} Posts 
H_{i} is the partial sum.
H_{1} = 1 H_{2} = 3/2 H_{3} = 11/6 H_{4} = 25/12 H_{5} = 137/60 For n=5 you want to show 1 + 3/2 + 11/6 + 25/12 + 137/60 = 6*(137/60)  5 This is true, at least for n=5, so there is a hope of finishing the proof. This is the Homework Help forum, so I think we've done our help by clarifying the problem statement  it's time for you to try some more. If you need more help, please come back with a summary of what you've tried and where you are stuck. 
20100129, 01:25  #9 
Sep 2004
13·41 Posts 
i got it! now i'm doing a similar one with fibinochhi numbers

20100129, 08:23  #10 
Dec 2008
2·5·83 Posts 
Never heard of "fibinochhi" numbers before, but I have heard of Fibonacci numbers
Last fiddled with by flouran on 20100129 at 08:23 
20100129, 10:35  #11 
Sep 2004
1000010101_{2} Posts 
yah i knew my chance of spelling it right was low, but I got it down anyway. My spelling is bad after midnight.

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Polynomial Algebra  R.D. Silverman  Other Mathematical Topics  15  20141029 19:42 
QS  lin. algebra phase  SilverMan  Factoring  39  20091217 16:41 
Linear algebra at 600%  CRGreathouse  Msieve  8  20090805 07:25 
Algebra Problem  davar55  Puzzles  15  20060501 16:55 
abstract algebra  DSC  Miscellaneous Math  2  20051004 09:14 