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enter image description here Here the series indeed reduces to a logarithmic value. How does it come to that .please explain

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    While obviously not what you wanted, I've updated my answer so that it may be done without a calculator. – 2017-01-21

4 Answers 4

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As $$\frac{1}{3n+k}=\frac{1}{n}\cdot\frac{1}{3+\frac{k}{n}},$$ the sum you are asking for, approximates the integral $$\int_0^1\frac{\text{d}x}{3+x}=\ln\frac{4}{3}.$$

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It suffices to see that

$$\frac14=\frac n{4n}=\sum_{k=1}^n\frac1{3n+n}\le\sum_{k=1}^n\frac1{3n+k}\le\sum_{k=1}^n\frac1{3n+0}=\frac n{3n}=\frac13$$

Thus, we see that $(a)$ is the correct answer, as it is the only one between $\frac14$ and $\frac13$.

Note that it is easy enough to check without a calculator, since $\ln(a/b)=\int_b^a\frac1x\ dx$

$(a):$

$$\frac14=\int_3^4\frac14\ dx<\int_3^4\frac1x\ dx<\int_3^4\frac13\ dx=\frac13$$

$(b):$

$$\ln(3/4)<\ln(1)<0<\frac14$$

$(c):$

$$\frac13=\int_2^3\frac13\ dx<\int_2^3\frac1x\ dx<\int_2^3\frac12\ dx=\frac12$$

$(d):$

$$\frac15=\int_4^5\frac15\ dx<\int_4^5\frac1x\ dx<\int_4^5\frac14\ dx=\frac14$$

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    :D I love multiple choice questions... – 2017-01-21
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    @projectilemotion Haha, I do that all the time. Almost wish they could raise the daily limit... – 2017-01-21
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    But why the sequence converges? The above answers involving the integral answet this question. – 2017-01-21
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    @szw1710 You can see the other answers or you could note that being bounded tightly enough is all that's needed to answer a multiple-choice question. (there is no "it diverges" as an option...) – 2017-01-21
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    If this is multiple-choice question (this really is), then you are right. In fact I like your answer, which is reasonable. Nevertheless, a true mathematician would like to know why... :-) You have my upvote. – 2017-01-21
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It is a well-known trick:

$$ \sum_{k=1}^{n} \frac{1}{3n+k} = \sum_{k=1}^{n} \frac{1}{3+(k/n)} \cdot \frac{1}{n} \xrightarrow[n\to\infty]{} \int_{0}^{1} \frac{dx}{3+x} = \log(4/3). $$

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Using $\:\displaystyle\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^nf\left(\frac{k}{n}\right)=\int_0^1f(x)\;dx$, we get:

$$\lim_{n\to \infty}\sum_{k=1}^n\frac{1}{3n+k}=\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\frac{1}{3+\dfrac{k}{n}}=\int_0^1\frac{dx}{3+x}=\ldots=\log\frac{4}{3}.$$

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    I was looking for this formula through which all answer came up. – 2017-01-21