monotonic sequence, is it increasing and is it bounded? is my work right?

an=n+1/n (btw anything that i write with a less than sign i mean less than or equal to i just cant write it.)

so i figured out that

a1=2

a2=2 1/2

a3=3 1/3

so my guess is that its increasing.

so for n=k

i need to show ak<ak+1

a1=2<ak=2 1/2 so i showed this right i believe.

next, i need to show that ak+1<ak+2

so i did (k+1)+(1/k+1)<(k+2)+(1/k+2)

cancelled out and reduced to (k+1)+1<(k+2)+1

reduced to (k+1)<(k+2)

reduced to 1<2

is this right to show that it is increasing?

then i just took the limit of n+1/n and got infinity, so it is not bounded.

is this all right? my work and all? my professor did it a different way. Thank you guys!

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

Quote:

Originally Posted by

**abowler10154** so i did (k+1)+(1/k+1)<(k+2)+(1/k+2)

cancelled out and reduced to (k+1)+1<(k+2)+1

And what did you cancel out? And please don't say 1/k.

If $\displaystyle a_n=1+1/n$, you don't need to use the induction hypothesis, so you don't need to use induction at all. However, I am not sure the definition is not $\displaystyle a_n=(n+1)/n$.

In plain text, it is customary to write a_n for $\displaystyle a_n$ and <= for $\displaystyle \le$.

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

I made common denominators on each side, k+1 being the left common one and k+2 being the right one so i could add them individually and cancel if you understand what i mean. But is the way i did it correct also? or no.

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

Actually, thinking about it, i dont believe i can cancel...i think im forgetting the rules of 8th grade math in college now. So if i got to the point where i cancelled is that okay? then how would i proceed if so?

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

It seems to me that you omitted parentheses around k+1 in 1/k+1 and treated it as a sum instead of a fraction that it is.

If $\displaystyle n>1$, then $\displaystyle n+1+\frac{1}{n+1}>n+1>n+1/n$. The last inequality holds because 1/n < 1. No use of induction hypothesis, and thus of induction, is necessary.

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

Quote:

Originally Posted by

**abowler10154** an=n+1/n (btw anything that i write with a less than sign i mean less than or equal to i just cant write it.)

For this kind of sequence think of the function $\displaystyle f(x)=x+\tfrac{1}{x}$.

Now $\displaystyle f'(x)=1-\tfrac{1}{x^2}$ which tells us its increasing. But looking it, it is not bounded for $\displaystyle x>1$.

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

I forgot that i can take the derivitive and if its positive its increasing. Thank you, so now how do i show the work that it is not bounded? Its obviously bounded below as a1 is 2 and its increasing. but what about being bounded above?

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

Quote:

Originally Posted by

**abowler10154** I forgot that i can take the derivitive and if its positive its increasing. Thank you, so now how do i show the work that it is not bounded? Its obviously bounded below as a1 is 2 and its increasing. but what about being bounded above?

Clearly it is not bounded above.

Re: monotonic sequence, is it increasing and is it bounded? is my work right?

ok, another way to show its increasing...how about this?

an<an+1

(n^2+1)/n<((n+1)^2+1)/(n+1)

cross multiply and i get that 2<4. does this prove its increasing?