mjhjmtu

An easy way to visualize why this can't be true is to try putting some points on a number line.

Start with 1 point in [0, 1):

2 points in [1, 2):

And so on:

Now you have a sequence that grows to infinity but keeps getting closer together.

Any increasing sequence $a_n_ngeq 1$ has limit in $mathbbRcup+infty$. It is $sup_ngeq 1 a_n$. Such $sup$ or supremum can be a finite number or $+infty$ (even if we know that $a_n+1-a_nto 0$).

An example with a finite limit is $a_n=1-1/nto 1$ and $a_n+1-a_n=frac1n(n+1)to 0$.

On the other hand $a_n=sqrtnto +infty$ and $a_n+1-a_n=sqrtn+1-sqrtn=frac1sqrtn+1+sqrtnto 0$.

So, the answer is NO, the condition $a_n+1-a_nto 0$ is not sufficient for an increasing sequence $a_n_ngeq 1$ to have a FINITE limit.

Another counterexample is $a_n=ln n$, for $ngeq1$. The difference of successive terms is $ln(n+1)-ln n = ln (1+1/n) rightarrow ln 1 = 0$, as $n rightarrow infty$, yet $ln n$ itself tends to infinity, as $n$ tends to infinity.

No. Consider the sequence $a_n_n=1^infty$ given by

• 
  $a_n = sumlimits_k=1^n frac1k$.

It follows that

• $a_n > a_n-1$


• 
  $a_n - a_n-1 = frac1n rightarrow 0$ as $n rightarrow infty$, but


• 
  $a_n = sumlimits_k=1^n frac1k rightarrow infty$ as $n rightarrow infty$ (by, e.g., integral test).

The condition $a_n+1-a_n to 0$ is not sufficient, as José Carlos Santos pointed out. But, a necessary and sufficient condition, that doesn't require the series to be increasing, is that $limlimits_ntoinfty(a_n+m(n)-a_n)=0$ for all $m(n)in mathbbN$, where $m$ is a function of $n$. Sequences which satisfy this property are called Cauchy sequences.

Also, if you show that a sequence is monotonically increasing and bounded from above, then it converges. The same applies for monotonically decreasing sequences which are bounded from below.

Note that if we define $b_n=a_n+1-a_n$, then $a_n=a_0+sum_n=0^inftyb_n$. So this question is equivalent to asking whether the terms of an infinite series going to zero is sufficient for the series to converge. There are a variety of examples of series with terms that go to zero, yet do not converge, with the harmonic series ($sum frac 1 n$) being one of the most famous.

And in fact we can construct a counterexample from any sequence by defining a sequence $c_n$ by simply re-indexing the terms. We set $c_0$ equal to $a_0$. Then set $c_k1$ equal to $a_1$, where $k_1>a_1-a_0$, and fill in the terms $c_1$ to $c_k-1$ with equally spaced terms; this will result in all of the consecutive differences from $c_0$ to $c_k1$ being less than $1$. Then set $c_k2$ equal to $a_2$, where $k_2+k_1>2(a_2-a_1)$, which results in consecutive differences between $c_k1$ to $c_k2$ being less than $frac 1 2$. Just keep re-indexing each term and filling in more and more new terms, and you can drive the consecutive differences arbitrarily low.

Another approach is to look at a sequence as an approximation of a continuous function, and the difference between successive terms as an approximation of the derivative. Then we just need a function such that $f'(x)$ converges to zero, but $f$ diverges. Two examples of such are the log function, (which gives a sequence very similar to the harmonic sequence) and square root. Note that both of these examples can be obtained by taking the inverse of a function whose derivative is constantly increasing. If $g'$ goes to infinity, then $(g^-1)'$ goes to zero. But if the domain of $g$ is the whole real line, then the range of $g^-1$ is the whole real line, i.e. $g^-1$ goes to infinity.