Proof of divergence of 1/2 + 1/(2 + 1/2) + 1/(3 + 1/3) + ...

countabl...@gmail.com

unread,

May 17, 2009, 4:42:19 PM5/17/09

to

Consider a series where the nth term is 1/(n + 1/n). How can we prove
that this series diverges?

Patrick Coilland

unread,

May 17, 2009, 4:59:52 PM5/17/09

to

countabl...@gmail.com a �crit :

> Consider a series where the nth term is 1/(n + 1/n). How can we prove
> that this series diverges?

1/(n+1/n) > 1/(n+1) hence the result

veal...@guerillamail.org

unread,

May 17, 2009, 5:03:42 PM5/17/09

to

On May 17, 4:42 pm, countableinfin...@gmail.com wrote:
> Consider a series where the nth term is 1/(n + 1/n). How can we prove
> that this series diverges?

I chose to compare it with the divergent sum of reciprocals: 1/2 + 1/3
+ 1/4 + ...
Call the series in which the nth term is 1/(n+1/n) S.
Rewrite each term of S as n/(n^2+1).
n/(n^2+1) = 1/(n+1) + n/(n^2+1) - 1/(n+1)
= 1/(n+1) + (n-1)/((n^2+1)(n+1))
with the exception of the first term, each term of S is greater than
the respective term in the reciprocal series.
Therefore, S must be divergent.

Martin Michael Musatov

unread,

May 17, 2009, 5:23:47 PM5/17/09

to

http://groups.google.com/group/sci.math/browse_thread/thread/16d3720a3cffc651/6cf04dc7e0b02403?q=P%3DNP

Thank you for the supporting evidence.

Martin Musatov [P==NP]

Martin Musatov

unread,

May 17, 2009, 5:31:26 PM5/17/09

to

http://mathforum.org/kb/message.jspa?messageID=6714174&tstart=0
Or convergent. {s}http://groups.google.com/group/Atheism-vs-
Christianity/browse_thread/thread/88a32aff562086d9#
Don't look at me, I am just pointing out two logical conditions.

Martin Musatov

unread,

May 17, 2009, 5:32:52 PM5/17/09

to

I am a Christian, I don't even know Hebrew, I just read about online.
http://groups.google.com/group/Atheism-vs-Christianity/browse_thread/thread/88a32aff562086d9#

Gerry Myerson

unread,

May 17, 2009, 6:47:00 PM5/17/09

to

In article <4a107ad0$0$12613$ba4a...@news.orange.fr>,
Patrick Coilland <pcoi...@pcc.fr> wrote:

That's probably the best answer. Alternatively, there's
the integral test, since 1 / (x + (1 / x)) = x / (x^2 + 1) is easy
to integrate.

--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

Martin Musatov

unread,

May 17, 2009, 10:31:33 PM5/17/09

to

On May 17, 3:47 pm, Gerry Myerson <ge...@maths.mq.edi.ai.i2u4email>
wrote:
> In article <4a107ad0$0$12613$ba4ac...@news.orange.fr>,
> Patrick Coilland <pcoill...@pcc.fr> wrote:
>
> > countableinfin...@gmail.com a écrit :

> > > Consider a series where the nth term is 1/(n + 1/n). How can we prove
> > > that this series diverges?
>
> > 1/(n+1/n) > 1/(n+1) hence the result
>
> That's probably the best answer. Alternatively, there's
> the integral test, since 1 / (x + (1 / x)) = x / (x^2 + 1) is easy
> to integrate.
>
> --
> Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)

Powered by infinity: ∞