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Consider the sum of $n$ terms :

$S_n = 1 + \frac{1}{1+2} + \frac {1}{1+2+3} + ... + \frac {1}{1+2+3+...+n}$ for $n \in N$.

Find the least rational number $r$ such that $S_n < r$, for all $n \in N$.

My attempt :

$S_n = 2(1-\frac{1}{2} + \frac {1}{2} - \frac{1}{3} + .... + \frac {1}{n} - \frac {1}{n+1}) = 2(1 - \frac {1}{n+1}) $

Now what to do with that '$r$' thing ?

How to proceed ?

  • 0
    Do you know what limits are? – 2017-01-13
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    @Nirbhay. Can you tell what is $r$ – 2017-01-13
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    hint: $S_n=\sum_{i=1}^{n}{\frac{2}{i(i+1)}}=2\sum_{i=1}^{n}\bigg( {\frac{1}{i}-\frac{1}{i+1}}\bigg) $ – 2017-01-13
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    @H.Ergül That's exactly what the OP wrote from the beginning... – 2017-01-13
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    Try making your titles informative whenever you can fit your main question in the title in its entirety. – 2017-01-13
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    It looks like you have an off-by-one mistake in your sum formula. Comparing with what @H.Ergül wrote, it should be $S_n=2(1-\frac1{n+1})$, as can be easily verified for some small $n$. – 2017-01-14
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    @MvG Thanks for telling, I will edit accordingly... – 2017-01-14
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    [This question](http://math.stackexchange.com/questions/2022169/how-to-prove-using-induction-that-sum-i-1n-frac1-sum-n-0i-n) deals with showing $S_n=\frac{2n}{n+1}$. – 2017-01-15

2 Answers 2

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Let $r=2$, and we can see that

$$\frac{2n-2}n=2-\frac2n<2$$

Similarly, as $n\to\infty$, the limit is $2$, so this is the least rational number satisfying the inequality.

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First, let's point out that it isn't obvious why there is a "least" such rational number. For instance, if $S_n = 2$ for all $n$, then there isn't such a least rational number. So let's first prove that $r$ exists.

Let $A = \{x \in \Bbb Q: S_n < x, \forall n\}$. The question should read:

Prove that $\min A$ exists and find its value.

$$S_n = \sum_{k=1}^n \frac1{\sum_{t = 1}^k t} = \sum_{k=1}^n \frac1{\frac{k(k+1)}2} = 2\left( \sum_{k=1}^n \frac1k - \sum_{k=1}^n \frac1{k+1}\right) \\ = 2 \left(1 - \frac1{n+1} \right)$$

If $x \in A$, then $S_n < x, \forall n \implies \lim S_n \le x \implies 2 \le x$, so $2$ is a lower bound of $A$. On the other hand, $2 \in A$. Thus, $2 = \min A$; in particular, $\min A$ exists.

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    While your overall point is well-taken, the question as stated is not wrong. Effectively the questioner is saying "It can be proved that there is a least rational number $r$ such that $S_n < r$ for all $n$, but we are not asking you to do that. We just want you to find the number $r$". As long as it is true that such an $r$ exists, the question is acceptable. – 2017-01-13