Time invariance of a summation sequence The following sequence is given which is supposed to be time-variant:
$$y[n] = \sum_{k=n_0}^n x[k]$$
I'm having difficulties proving the time-variance or finding a counterexample for it being time-invariant.
My idea (which proves it being time-invariant?) is:
\begin{align}y_1[n] &amp;= T\{x_1[n]\}\\x_2[n] &amp;= x_1[n-n_0]\end{align}
check, whether    $T\{x_1[n]\} = y_1[n-n_0]$
$$y_2[k] = \sum_{k=n_0}^n x_2[k] = \sum_{k=n_0}^n x_1[k - n_0] = \sum_{k=n_0-n_0}^{n-n_0} x_1[k] = y_1[n-n_0]$$
Where is my mistake? Working around the summation is giving me trouble.
EDIT:
Following the advice in the comments for $n = n_0$ the left equation equals to 1 while the right equation equals to 0, making both equations unequal and therefore the sequence time-variant?
Second sequence
This sequence is supposed to be time-invariant.
\begin{align}y[n] &amp;= \sum_{k=n-n_0}^{n+n_0} x[k]\\
y_2[k] &amp;= \sum_{k=n-n_0}^{n+n_0} x_2[k] = \sum_{k=n-n_0}^{n+n_0} x_1[k - n_0] = \sum_{k=n_0-2n_0}^{n} x_1[k] = y_1[n-n_0]\end{align}
I followed the same steps but I'm still not sure about for the case $n = n_0$ since when added in left and right equation
$$\sum_{k=0}^{2n} = \sum_{k=-n}^{n}$$
They seem unequal to me?
Sequences that are &quot;time-invariant&quot; must be constants (not varying with time at all).
What the OP is actually asking about is a (discrete-time) system whose response $y$ to the input $x$ is defined to be
$$y[n] = \begin{cases}\sum_{k=n_0}^n x[k], &amp; n \geq n_0,\\??? &amp; n &lt; n_0\end{cases}$$
where the $???$ is there because according to some traditionalists, if $n &lt; n_0$, then that sum is an empty sum which has value $0$ but more liberal-minded might treat that sum as representing $$x[n] + x[n+1] + x[n+2] + \cdots + x[n_0-1] + x[n_0]$$ (with more serious-minded folks even negating that sum in analogy with the standard result that for $a&lt;b$, $\int_b^a = -\int_a^b$). So, for the traditional meaning of things and $n_0$ denoting a (finite) integer, the system is not a time-invariant system. It is time-invariant (with unit pulse response $h[n]$ being the unit step function $u[n]$) if $n_0 = -\infty$ as noted in a comment by Robert Bristow-Johnson on the main question.