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I'm having a hard time determining if the following series converges (absolutely?) or diverges:

$$\sum_{i=1}^n \frac 1 {2+\sin n}$$

I would really appreciate some help here. Thanks!

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  • $\begingroup$ The series diverges by the Comparison test. $\endgroup$ – Idonknow Nov 15 '15 at 3:15
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    $\begingroup$ Are you sure you mean sin(n) and not sin(i)? Either way you can use comparison to show divergence $\endgroup$ – Ryan Nov 15 '15 at 3:50
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    $\begingroup$ How can there be any convergence or divergence if there is no limit being taken? Your question is unclear. $\endgroup$ – Ali Caglayan Nov 15 '15 at 9:12
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Note that we have $$0 \leq \dfrac{1}{2 + 1} \leq \frac{1}{2+\sin(k)}$$

for all $k \in \mathbb{N}$.

Since $\sum_{k \in \mathbb{N}} {\dfrac{1}{3}}$ diverges, by the Comparison test, the sum $$\sum_{k \in \mathbb{N}}\dfrac{1}{2+ \sin(k)}$$

diverges.

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The "infinite" sum diverges ($\lim_{n \rightarrow \infty} c_n \not = 0$) . But your finite sum should converge, just add all the terms together up to n.

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  • $\begingroup$ Won't all sums of finite length converge (as long as all elements are finite, of course)? $\endgroup$ – Mico Nov 15 '15 at 8:53
  • $\begingroup$ Yes. As long as they are finite, add up term by term will give convergence. $\endgroup$ – Paichu Nov 15 '15 at 8:54
  • $\begingroup$ But is convergence even an interesting or useful concept for finite-length sequences and their sums? $\endgroup$ – Mico Nov 15 '15 at 8:57
  • $\begingroup$ It certainly does. For example algebraic manipulation on sum can be very strange if doing it on infinite sum (for example, sum of all natural number is $-\frac{1}{12}$). In applied fields, I believe finite sums are used more often. $\endgroup$ – Paichu Nov 15 '15 at 9:05
  • $\begingroup$ My comments are about the usefulness of the convergence concept for finite sums, yet your reply asserts the usefulness of the concept for infinite sums. I think we're talking past each other $\endgroup$ – Mico Nov 15 '15 at 9:46
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Observe that $\lim_{n\rightarrow \infty} \frac1{2+\sin n}\neq0$ Hence the series diverges.

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The terms do not approach $0$ since $1 \le 2+\sin n \le 3$, so $\dfrac 1 3 \le \dfrac 1 {2+\sin n}$. Therefore the series diverges.

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