Showing that $S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000} \gt 1$

Question

If $$S=\frac{1}{100} + \frac{1}{101} + \dots + \frac{1}{1000}$$ then $$S\gt 1,$$

but how?

I understood that there are $451$ pair of terms. So clubbed two terms together. $\frac{1}{100}+\frac{1}{1000}+\frac{1}{101}+\frac{1}{999}.....$ But I am not able to solve it further. Such a tricky question for me.

Answer 1

Hint

$$\frac{1}{100}+\dots+\frac{1}{499}> \frac{400}{500}=\frac{4}{5}.$$

Answer 2

There are $901$ terms in $S$

Consider the first $100.$ The smallest term of this set is $\frac 1{199}$

The sum of the first $100$ terms is greater than $\frac 12$

Now consider the next $200$ terms.

the smallest term in this set is $\frac {1}{399}$

and the sum of that subset is greater that $\frac 12$

and you are done. And there are still $601$ you haven't even considered yet.

The harmonic series:

$S = \sum_\limits{n=1}^\infty \frac 1n = 1 + \frac 12 + \frac 13 + \frac 14 \cdots\\ \frac 13 + \frac 14 > \frac 1{4} + \frac 1{4} = \frac 12\\ \frac 15 + \frac 16 + \frac 17 + \frac 18 > \frac 18 + \frac 18 + \frac 18 + \frac 18 = \frac 12\\ $

And we can break infinite series into (infintely many) finite sub-series, and each sub-series sums to something that is greater than $\frac 12$

$S > 1+\frac 12 + \frac 12 + \frac 12 \cdots$ and as we add more terms $S$ marches off to infinity (albeit slowly)

For you problem I have employed a very similar tactic. I have found subsets of your series, each of which sum to something greater than $\frac 12$

Answer 3

\begin{eqnarray*} \frac{1}{129}+\cdots + \frac{1}{256} > \frac{1}{2} \\ \frac{1}{257}+\cdots + \frac{1}{512} > \frac{1}{2} \\ \end{eqnarray*}

Answer 4

If you draw a graph, the following inequality should be clear:

$$\sum_{n=100}^{1000} \frac{1}{n} > \int_{100}^{1001} \frac{1}{x} \; dx = \ln 1001-\ln 100 = \ln \frac{1001}{100} >\ln 10. $$

And since $10>e$, the last is greater than $1$.

Answer 5

By C-S $$\frac{1}{100}+\frac{1}{101}+...+\frac{1}{1000}\geq\frac{901^2}{100+101+...+1000}=\frac{901^2}{\frac{100+1000}{2}\cdot901}=\frac{901}{550}>1$$

Answer 6

numbers from $\frac{1}{100}$ to $\frac{1}{200}$ are larger than $\frac{1}{200}$

and there are $100$ of them (101 actually, but doesn't matter) so their sum is larger than $100\times \frac{1}{200}=\frac12$

You can repeat a similar argument to show that from $\frac{1}{200}$ to $\frac{1}{300}$ the sum is larger than $\frac13$

until you arrive to numbers from $\frac{1}{900}$ to $\frac{1}{1000}$ which have sum larger than $\frac{1}{10}$

so the numbers from $\frac{1}{100}$ to $\frac{1}{1000}$ are less than $\frac12+\frac13+\ldots+\frac{1}{10}\approx 1.93 $

Our sum is less than the actual sum ($\approx 2.308$) and is larger than $1$ so we have proved what was required to prove

hope this helps