1. Guess the supremum and infimum of the set $A=\left\{\left(\frac{n+1}{2n}\right) \left(1+\frac{1}{n}\right)|n\in\mathbb{N}\right\}$ and prove by definition that they are indeed.

  2. Prove by definition that the $$\lim_{n\to \infty}~~ \left(\frac{n+1}{2n}\right)\left(1+\frac{1}{n} \right)=\frac{1}{2}$$


  1. My guess for $\inf$ and $\sup$ are $\inf(A)= 1/2$, $\sup(A)=2$. Although I'm not sure how to prove this by definition... Must I prove this by showing they are lower bounds and upper bounds respectively and then show that they are the smallest/greatest upper bound ? How would I go about doing this ?

  2. I am quite certain that this isn't a valid proof since I haven't proved anything, it just works... but:

Let $A_n$ denote the $n^{th}$ term in the set A and consider $N=\frac{1}{\epsilon}$.

Then $\forall\epsilon>0,~~\left|A_n-\frac{1}{2}\right|<\epsilon~~~\forall n>N$

How could I make this more formal?

  • $\begingroup$ Your claim that for $N=\frac{1}{\epsilon}$, then for all $\epsilon>0$ we have $\left|A_n-\frac{1}{2}\right|<\epsilon$ for all $n>N$ is false. Try $\epsilon=2$ and $n=1$. $\endgroup$
    – user940
    Oct 12, 2015 at 22:55
  • $\begingroup$ Ok good point. How may I prove it then? $\endgroup$ Oct 13, 2015 at 10:07

1 Answer 1


Hint: Both $1+\frac1n$ and $\frac{n+1}{2n}$ are decreasing.

  • $\begingroup$ Ok that helps me to prove the Sup(A) = 2 but how does that help me prove the Inf(A)= 1/2 ? $\endgroup$ Oct 13, 2015 at 10:06
  • $\begingroup$ @Pondzosaurusrex It is easy to verify, that all elements are greater than $1/2$. Proving that it is infimum is in this case the same as proving that limit is equal to $1/2$. Or: or every $\epsilon>0$ there is $n_0$ such that $1/n<\sqrt{\epsilon}$ (and so $1/2n$) for all $n\geq n_0$. Do you see the end of the proof? $\endgroup$
    – Kola B.
    Oct 13, 2015 at 21:23

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