Monotonia di successione con tangente

[keliaxv]

Buongiorno potreste aiutarmi a studiare la monotonia di questa successione?
$a(n)= 1/n*tan(1/n)$
ho provato a studiare $1/(n+1)*tan(1/(n+1))-1/ntan(1/n)>0$ ma non riesco a trovare i valori in cui è positiva... c'è un altro modo che mi sfugge o dovrei riuscire a risolvere la disequazione?

[javicemarpe]

Try to treat it as a function in a neighbourhood of the origin. How would you check if it is increasing or decreasing?

[keliaxv]

you mean doing the limit $lim_(x->0)(1/xtan(1/x))$
but it doesn't exist

[javicemarpe]

No, you should thinks of it as $x tan(x)$ when $x$ is close to the origin ($x$ close to zero will do the rol of $\frac{1}{n}$, because $\frac{1}{n}$ is close to the origin)

[keliaxv]

ok, so studying $lim_(x->0)(xtan(x))$ is like studying $a_n$ for $n_{n \to \infty}$ so I see that it's infinitesimal for $n_{n \to \infty}$... but is it enough to say that $a_n$ is monotonic?

[javicemarpe]

No, you don't have to study that limit, but the function $x tan(x)$ when $x$ is close to $0$. If this function is increasing (or decreasing) in a neighbourhood of $0$ sufficiently large, then your sequence will also be increasing (or decreasing) because its points are points on the graph of $x tan(x)$. The advantage of this approach is that now you can use differential calculus.

[keliaxv]

Ah ok, you're saying that increasing n, after n=1, my $a_n$ is similar to $xtan(x)$ with x close to 0? Of course the function is decreasing for x->0 and so does my sequence (not sure about this term) in its domain?

[javicemarpe]

As $x \tan x$ is an increasing function when $x$ is close to the origin, you have that your sequence is decreasing (because its points go to zero when $n$ increases) at least when $n$ is large enough.