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Find the sum of this infinite series: $$1+\frac 12+\frac 13+\frac14+\ldots \infty=?$$

$a)\ \pi\quad b)\ \pi^2/4\quad c)\ 4\quad d)\ \infty $

my try: $$1+\frac 12+\frac 13+\frac14+\ldots \infty=\sum_{r=0}^n\frac{1}{n+r}$$ $$=\int_{0}^1\frac{1/n}{1+/nr}$$$$=\int_{0}^1\frac{dx}{1+x}=\ln(x)_0^1=\ln2$$ I think my answer is wrong. I don't know how to reach the answer. someone please help me.

Thanks

  • 4
    this infinity sum is infinity – 2017-01-29
  • 10
    I am _so_ happy that $-1/12$ was not one of the given options. Anyways, the standard proof of this is to compare your series with $$1+\frac12+\frac14+\frac14+\frac18+\frac18+\frac18+\frac18+\frac1{16}+\cdots$$ – 2017-01-29
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    Really? 1k rep and you do not know the harmonic series is divergent? – 2017-01-29
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    https://en.m.wikipedia.org/wiki/Harmonic_series_(mathematics) – 2017-01-29
  • 10
    What's the 1k rep have to do with anything? Getting 1k rep on this site is relatively easy and certainly possible to do without any knowledge of series. – 2017-01-29
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    @Arthur You were so happy because you're not a physicist. :) – 2017-01-29
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    I don’t understand the “$\cdots+\infty$” just before the equals sign. – 2017-01-29
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    @pjs36 Oh I read it wrong, sorry! (I deleted my comment) – 2017-01-29
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    @AlanTuring No, I was so happy because we have enough questions about that damn thing already. I really like the "equality" personally, but I'm tired of trying to make people who have just discovered it realize that it is not gospel, or people who are asking "I thought the sum was infinite, how can it be infinite when it's $-1/12$? Is $-1/12 = \infty$?" or any other variation. – 2017-01-29
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    Why is that question closed ? – 2017-01-31
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    @Lubin. The $\infty$ symbol before the equals sign shouldn't be there. – 2017-01-31

4 Answers 4

7

This equality is false: $$1+\frac 12+\frac 13+\frac14+\ldots \infty=\sum_{r=0}^n\frac{1}{n+r}$$

First, the $\infty$ shouldn't be there on the LHS. But that's not why the equality is false. The upper limit on the $\sum$ should be $\infty$, not $n$. Also, no need to complicate things by introducing the $n$ at all. Replacing $n$ with $1$ will make it correct but it's better to remove it entirely and just start the summation at $r=1$. So the equality really is $$1+\frac 12+\frac 13+\frac14+\cdots=\sum_{r=1}^\infty\frac1r$$

This sort of invalidates all the steps after that, which didn't make much sense anyway (but to be fair I didn't spend a lot of time analyzing them).

Anyway, this is the well-known harmonic series, which diverges, so the answer is D. See Arthur's comment on your question for a proof idea.

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    The RHS of the equality does converge to $\ln 2$ but it's not the same thing as thr LHS.............+1 – 2017-01-31
3

That is the harmonic series

$$\sum_{k = 1}^{+\infty} \frac{1}{k}$$

You can test the divergence with the integral test:

$$1 + \frac{1}{2} + \frac{1}{3} + \ldots = \int_0^1 (1 + x + x^2 + x^3 + \ldots)\text{d} x$$

This is equal to

$$\int_0^1\left(\sum_{K = 0}^{+\infty} x^K\right)\ \text{d}x = \int_0^1 \frac{1}{1-x}\ \text{d}x = +\infty $$

3

A straightforward way of showing the divergence of this series is the following:

Compare it with the series $$ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{16} ... $$

In this series, there are $ 2^{k-1} $ terms of the form $ \frac{1}{2^k} $, and each term is less than or equal to the corresponding term for the harmonic series, so by the comparison test, if this series diverges, then the harmonic series must also diverge.

The terms can be grouped together so that when the $ 2^{k-1} $ terms of the form $ \frac{1}{2^k} $, each one gives a term $ \frac{1}{2} $, so the series becomes

$$ 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + ... $$

Since the limit of these terms is not zero, the series cannot converge, and since this series is less than or equal to the harmonic series in comparison, the harmonic series must diverge as well.

1

This is a generalized harmonic series: that is

$$\sum_{n=1}^\infty \frac{1}{n^\alpha};$$

in particular this serie converge iff $\alpha >1$. Check here for a proof https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Since in your case $\alpha=1$ the sum is infinity.

  • 2
    I don't really think this is a great answer. OP wants to know how to find sum of series (if it exists); this is only addressed by linking to a Wikipedia article: it is not self-contained. (Also putting the cart before the horse to say this is a specific example of a generalized harmonic series; this is the one that started the whole thing!!) – 2017-01-29