For any value of $n\ge5$, the value of $1+ \frac{1}{2}+ \frac{1}{3}+...+\frac{1}{2^n+1}$ lies between

Question

For any value of $n\ge5$, the value of $$1+ \frac{1}{2}+ \frac{1}{3}+...+\frac{1}{2^n+1}$$ lies between

A) $0$ and $\frac{n}{2}$;

B) $\frac{n}{2}$ and $n$;

C) $n$ and $2n$;

D) none of them

I know this is Harmonic progression, but how exactly am I supposed to know the interval?

Try to compare the sum with something you can derive to closed form. — M.P, May 09 '17 at 08:14
Calculate the answer for $n=5$ and see the answer is around $4$. We can see that $4<n=5$ and $4>\frac n2 = 2.5$, so we can discount A and C immediately. You then just need to prove if it is still between $\frac n2$ and $n$ when $n>5$ — lioness99a, May 09 '17 at 08:17
Also note that this sequence can be written as $$\sum_{k=1}^{2^n+1}\frac 1k$$ Does that help? — lioness99a, May 09 '17 at 08:25
Thanks it helps a lot! @lioness99a only if I could give you point for this ^_^ — Iti Shree, May 09 '17 at 08:26
$H_n$ behaves like $\log n$, hence your sum is approximately $n\log 2$. — Jack D'Aurizio, May 09 '17 at 12:45
A similar question: Prove this inequality: $\frac n2 \le \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac1{2^n - 1} \le n$ — Martin Sleziak, Aug 24 '19 at 06:01

score 4 · Accepted Answer · answered May 09 '17 at 08:45

Note that $$\sum\limits_{k=1}^{2^n+1} \dfrac{1}{k} > \sum\limits_{k=2}^{2^n}\dfrac{1}{k} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots \dfrac{1}{2^n} > \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{8} + \dots + \dfrac{1}{2^n} = \sum\limits_{k=1}^n \dfrac{2^{k-1}}{2^k} = \dfrac{n}{2}$$

On the other hand, $$\sum\limits_{k=1}^{2^n+1} \dfrac{1}{k} \le \sum\limits_{k=1}^{2^{n+1} - 1} \dfrac{1}{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dots \dfrac{1}{2^{n+1}-1} < 1+ \dfrac{1}{2} + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{4} + \dots + \dfrac{1}{2^n} = \sum\limits_{k=0}^{n} \dfrac{2^k}{2^k} = n + 1$$

You can decrease the upper bound to $n$ because $\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4}-\dfrac{1}{5} + \dfrac{1}{4}-\dfrac{1}{6} \dots + \dfrac{1}{2^{n+1}-1}-\dfrac{1}{2^n} > 1$.

For any value of $n\ge5$, the value of $1+ \frac{1}{2}+ \frac{1}{3}+...+\frac{1}{2^n+1}$ lies between

1 Answers1